ON CERTAIN INEQUALITIES BETWEEN THE ZEROES, ALL ASSUMED TO BE REAL, OF A POLYNOMIAL AND THOSE OF ITS DERIVATIVE

Prof. TIBERIU POPOVICIU

1.
- —
  
  Consider a polynomial $f(n)$ of degree $x$ having all its real zeros and let
  (1)

x_{1}\leqq x_{2}\leqq\cdots\leqq x_{n}

these zeros.
We know that the derivative $f^{\prime}(x)$ also has all its real zeros. Let us denote by
(2)

y_{1}\leqq y_{2}<\cdots\leq y_{n-1}

the zeros of $f^{\prime}(x)$ .
Between the zeros (1) and (2) we first have the fundamental equality

\frac{1}{n}\sum_{i=1}^{n}x_{i}=\frac{1}{n-1}\sum_{i=1}^{n-1}y_{j}

expressing the equality of their respective arithmetic means. This equality is also true without the restriction of the reality of zeros.

There are also important inequality relations between the zeros (1) and (2). We have first

x_{i}\leqq y_{i}\leqq x_{i+1},\quad i=1,2,\ldots,n-1,

which are immediate consequences of Rolle's theorem. It should be noted that, as a result of the notations used, the
sign $=$ is valid in both formulas (4) if and only if $x_{1}=x_{i+1}$ .

LAGUERRE clarified the inequalities (4) by demonstrating that

\frac{(n-1)x_{1}+x_{i}+1}{n}\leqq y_{i}\leqq\frac{x_{i}+\left(n-1x_{i+1}\right.}{n}

(5)

^{but, as Mr. J v. Sz. NAGY 1} has shown ), we also have the more precise inequalities

\frac{(n-i)x_{i}+x_{i}+1}{n-i+1}\leqq\cdots\leqq\frac{x_{i}+ix_{i}+1}{i+1}

(6)

2.
- —
  
  We deduced property (6) from a general property of zeros $y_{j}$ depending on the zeros $x_{i}{}^{2}$ ). We will recall this property. The zeros $x_{i}$ can vary in any way on the real axis, but always respecting the convention (1) We can get an idea of this variation by imagining that the zeros $x_{i}$ are impervious material points (which does not exclude the possibility that two or more of these points are in the same geometric point of the real axis). Then the zeros $y_{i}$ are continuous functions of the $x_{i}$ . We can still reason in the following way. Let $X$ the coordinate point $x_{1},x_{2},\ldots,x_{n}$ in an ordinary space $n$ dimensions. The $x_{i}$ being subject to restriction (1), the point $X$ describes a certain region of this space. Or also $Y$ the coordinate point $y_{1},y_{2},\ldots,y_{n-1}$ in an ordinary space $n-1$ dimensions. So the point $Y$ is a continuous function of the point $X$ .

Let us complete the preceding observations with a very simple remark. The following (2) of $y_{j}$ can be called the sequence derived from the sequence (1) of $x_{i}$ . So if the $k$ first resp. the $k$ last points $x_{i}$ tend towards $-\infty$ resp. towards $+\infty$ , THE $k$ first resp. the $k$ last points $y_{j}$ also tend towards $-\infty$ resp. towards $+\infty$ . In addition, the points $y_{k+1},y_{k+2},\ldots,y_{n-1}$

⁰⁰footnotetext: 1. Julius V. Sz. NAGY „Über algebraische Gleichangen mit lauter reillen Wurzeln" Jahresb. d. D. Math. Ver., 27, 37-43, (1918). See also, R, GODEAU „On algebraic equations with all real roots* Mathesis, 45, 245=252, (1931). 2. Tiberru Popoviciu. „On algebraic equations with all real roots". Mathematica 9, 129-145, (1945).

resp. $y_{1},y_{2},\ldots,y_{n-k-1}$ tend towards the points of the derived sequence of $x_{k+1},x_{k+2},\ldots,x_{n}$ resp. $x_{1},x_{2},\ldots,x_{n-k}$ .

Now let's return to the property mentioned above. It is the following:

Lemma 1. - Zeros $y_{j}^{\prime}$ are non-decreasing functions of zeros $x_{i}$ .

To be complete and also to clarify the statement of the lemma we will give the proof. It is obviously sufficient to examine the variations of the $y_{i}$ when one of the points $x_{i}$ varies very little. Let's take zero $y_{i}$ and suppose that $x_{k}$ take a small positive increase $h$ . This requires $x_{k}<x_{k+1}\left(x_{n+1}=+\infty\right)$ . If $x_{i}=x_{i+1}$ And $k\neq i+1,y_{i}$ obviously does not vary. If $x_{i}=x_{i-1}$ And $k=i+1,y_{i}$ becomes larger according to the remark made above on inequalities (4). Now let $x_{i}<x_{i+1}$ and let's ask

		$\displaystyle f(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots\left(x-x_{n}\right)==\left(x-x_{k}\right)\varphi(x)$
		$\displaystyle g(x)=\left(x-x_{k}-h\right)\varphi(x)$

$h$ always being a fairly small positive number. To show that $y_{i}$ , has become larger it is enough to demonstrate that $g^{\prime}(x)$ has a zero in the open interval ( $y_{i},x_{i+1}$ .). In the left neighborhood of $x_{i+1}$ we have ³ )

\operatorname{sg}g^{\prime}(x)=(-)^{n-i-1}

(7)

We then have

g^{\prime}\left(y_{i}\right)=\varphi\left(v_{i}\right)+\left(y_{i}-x_{k}-h\right)\varphi^{\prime}\left(y_{i}\right)

But

\varphi\left(y_{i}\right)+\left(y_{i}-x_{k}\right)\varphi^{\prime}\left(y_{i}=0,\quad f\left(y_{i}\right)=\left(y_{i}-x_{k}\right)\varphi\left(y_{i}\right)\right.

g^{\prime}\left(y_{i}\right)=\frac{hf\left(y_{i}\right)}{\left(y_{i}-x_{k}\right)^{2}}

hence
3) We pose, as usual

\text{ sg }z=\left\{\begin{array}[]{r}1,\text{ si }z>0,\\ 0,\text{ si }z\geq 0,\\ -1,\text{ si }z<0.\end{array}\right.

(8)

\operatorname{sg}g^{\prime}\left(y_{i}\right)=\operatorname{sg}f\left(y_{i}\right)=(-1)^{n-1}

Equalities (7) and (8) demonstrate the property.
Finally, if one or more zesos $x_{i}$ start to grow, all zero $y_{i}$ between two $x_{i}$ different also begins to grow.
3. - Lemma 1 immediately allows us to delimit the sum

y_{i}+y_{i+1}+\cdots+y_{i+j-1},\quad 1\leqq i\leqq i+j-1\leqq n-1

(9)

To obtain a lower limit, it is sufficient to make the zeros decrease indefinitely. $x_{1},x_{2},\ldots,x_{i-1}$ and to make it decrease until $x_{i+j}$ the zetos $x_{i+j+1};x_{i+j+2},\ldots,x_{n}$ . From thêthè to obtain a higher limitation it is enough to indefinitely extend the zeros $x_{i+i+1},x_{i+j+2},\ldots,x_{n}$ and to increase the zeros $x_{1},x_{2},\ldots,x_{i-1}$ . until $x_{i}$ . The limitations sought are therefore given by the polynomials

\left(x-x_{i}\right)\left(x-x_{i+1}\right)\ldots\left(x-x_{i+j-1}\right)\left(x-x_{i+j}\right)^{n-i-j+1}

(10)

\left(x-x_{i}\right)^{y}\left(x-x_{i+1}\right)\left(x-x_{i+2}\right)\ldots\left(x-x_{i+j}\right)

(11)

and to obtain them it is enough to write equality (3) for these polynomials.

We thus obtain the following property expressed by
THEOREM 1. - Between the zeros (1) and (2) we have the following inequalities
(12) $y_{i}+y_{i+1}+\cdots+y_{i+j-1}\geq\frac{(n-i)\left(x_{i}+x_{i+1}+\cdots+x_{i+j-1}\right)+jx_{i+j}}{n-i+1}$
(13) $y_{i}+y_{i+1}+\cdots+y_{i+j-1}\leqq\frac{ix_{i}+(i+j-1)\left(x_{i+1}+x_{i+2}+\cdots+x_{i+j}\right)}{i+j}$

j=1,2,\ldots,n-i,\quad i=1,2,\ldots,n-1.

The proof of Lemma 1 also tells us the cases where in (12) and (13) we have equality. First, if $x_{i}=x_{i+j}$
we obviously had 10 worthy in (12) and in (13) since then

y_{i}=y_{i+j-1}=\cdots=y_{i+j-1}=x_{i}=x_{i+1}=\cdots=x_{i+j}

If $x_{i}<x_{i+j},i>1$ equality in (12) and if $x_{i}<x_{i+j},i+j<n$ the equality in (13) are impossible for a polynomial $f(x)$ of effective degree $n$ . If $i=1,x_{1}<x_{j+1}$ the equality in (12) only holds for the polynomial (10) and if $i+j=n,x_{n-j}<x_{n}$ the equality in (13) only holds for the polynomial (11).

For $i$ we find the inequalities (6). For $i=1$ , $j=n-1$ the two inequalities reduce to equality (3).

In particular, potur $i=1$ and for $i=n-i$ , we obtain the following property:

THEOREM 2. - The arithmetic mean of the first j zeros (2) is at most equal to the arithmetic mean of the $j+1$ or first zeros (1),

\frac{y_{1}+y_{2}+\cdots+y_{j}}{i}\leqq\frac{x_{1}+x_{2}+\cdots+x_{j+1}}{j+1}.

The arithmetic mean of the $j$ last (2) is at least equal to the arithmetic mean of $j+1$ last zeros (1),

\frac{y_{n-j}+y_{n-j+1}+\cdots+y_{n-1}}{j}\geq\frac{x_{n-j}+x_{n-j+1}+\cdots+x_{n}}{j+1}.

If we notice that by a linear transformation $x=\alpha x^{\prime}+\beta$ the derived sequence transforms into the derived sequence of the transform of the primitive sequence, we see that the lower limitations of the sums (9) can be deduced from their upper limitations and vice versa.
4. - By combining inequalities (12) and (13) (among which is also equality (3)), we can deduce other interesting limitations for the sums (9) Let us write

y_{i}+y_{i+1}+\cdots+y_{i+j-1}=\sum_{r=1}^{n-1}y_{r}-\sum_{r=1}^{i-1}y_{r}-\sum_{r=i+j}^{n-1}y_{r}

and let us take into account inequalities (12), (13) and equality (3).

By doing the calculations we find the following property THEOREM 3. - Between the zeros (1) and (2) we have the following inequalities

\frac{(n-i)\sum_{r=1}^{i}x_{r}+i(n-1)\sum_{r=i+1}^{i+j-1}x_{r}+i(i+j-1)x_{i+1}}{ni}\leqq

(14)

\leqq y_{i}+y_{i+1}+\cdots+y_{i+j-1}\leqq

(n-i-j+1)(n-i)x_{i}+(n-i-j+1)(n-1)\sum_{r=i+1}^{i+j-1}x_{r}+(i+j-1)\sum_{r-i+j}^{n}x_{r}

$(15)\leqq\quad n(n-i-j+1)$

j=1,2,\ldots,n-i,\quad i=1,2,\ldots,n-1

The equality in (14) is only possible for the polynomial

\left(x-\frac{1}{i}\sum_{r=1}^{i}x_{r}\right)^{i}\left(x-x_{i+1}\right)\left(x-x_{i+2}\right)\ldots\left(x-x_{i+j-1}\right)\left(x-x_{i+j}\right)^{n-i-j+1}

and in (15) is only possible for the polynomial

\left(x-x_{i}\right)^{i}\left(x-x_{i+1}\right)\left(x-x_{i+2}\right)\ldots\left(x-x_{i+j-1}\right)\left(x-\frac{1}{n-i-j+1}\sum_{r=i+j}^{n}x_{r}\right)^{n-i-j+1}

It is easy to see that in these cases equality actually takes place. The cases of equality easily result from what we said in the previous No.

The two limitations (12) and (14) of the sum (9) are distinct, unless $i=1$ when they are identical. It is enough to do successively

		$\displaystyle x_{1}=-1,x_{2}=x_{3}-\cdots=x_{i_{+j}}=0,$
		$\displaystyle x_{1}=x_{2}=\cdots=x_{i_{+j-1}}=0,\quad x_{i_{+j}}=1,$

to see that these two limitations do not result from each other.

We see in the same way that the upper limitations (13) and (15) are independent in the same sense. unless $i=n-j$ when they are identical.

For $j=1$ inequalities (14) and (15) become

\begin{gathered}\frac{(n-i)\sum_{r=1}^{i}x_{i}+i^{2}x_{i+1}}{ni}\leqq y_{i}\leqq\frac{(n-i)^{2}x_{i}+i\sum_{r=i+1}^{n}x_{i}}{n(n-i)}\\ i=1,2,\ldots,n-1\end{gathered}

which are also due to MJ v. Sz. NAGY ⁴ ).
We do not intend to study more fully the consequences of inequalities (12) and (13). In the following we will indicate another procedure for obtaining inequalities between the zeros (1) and (2).

$\S 2$ .

5.
- —
  
  We will first supplement Lemma 1 with another general property of zeros (1) and (2). Suppose that we replace the zeros $x_{k+1},x_{k+2},\ldots,x_{n}$ by their average arterial

\xi=\frac{x_{k+1}+x_{k+2}+\cdots x_{n}}{n-k}.

We then have $x_{k+1}\leqq\xi\leqq x_{n}$ and equalities are impossible if $x_{k+1}<x_{n}$ .

Let's ask

		$\displaystyle f(x)=\left(x-x_{k+1}\right)\left(x-x_{k+2}\right)\ldots\left(x-x_{n}\right)\varphi(x),$
		$\displaystyle g(x)=(x-\xi)^{n-k}\varphi(x).$

The derivative $g^{\prime}(x)$ has a single zero in each of the intervals $\left(x_{1},x_{2}\right),\left(x_{2},x_{3}\right),\ldots\left(x_{k-1},x_{k}\right),\left(x_{k},\xi\right)$ . We will demonstrate that the zero of $g^{\prime}(x)$ in the meantime $\left(x_{i},x_{i+1}\right)(i\leqq k)$ is greater than $y_{i}$ if $x_{i}<x_{i+1}$ .

Indeed, in the right neighborhood of $x_{i}$ We have

\operatorname{sg}^{\prime}(x)=(-1)^{n-i}

(16)

Noting that $\xi>y_{i}$ , or find first $\operatorname{sg}g^{\prime}\left(y_{i}\right)=(-1)^{n-k-1}\operatorname{sg}\left[(n-k)\varphi\left(y_{i}\right)+\left(y_{i}-\xi\right)\varphi^{\prime}\left(y_{i}\right)\right]$ .

But

4.

Loc. cit. 1).

\operatorname{sg}\varphi\left(y_{i}\right)=(-1)^{k-i},

\operatorname{sg}g^{\prime}\left(y_{i}\right)=(-1)^{n-i}\operatorname{sg}\left[\frac{\varphi^{\prime}\left(y_{i}\right)}{\varphi\left(y_{i}\right)}-\frac{n-k}{\xi-y_{i}}\right].

We have

\frac{\varphi^{\prime}\left(y_{i}\right)}{\varphi\left(y_{i}\right)}=\frac{1}{x_{k+1}=y_{i}}+\frac{1}{x_{k+2}=y_{i}}+\cdots+\frac{1}{x_{n}=y_{i}}

so, since $x_{k+1}-y_{i}>0$ ,

\frac{\varphi^{\prime}\left(y_{i}\right)}{\varphi}\frac{n-k}{\left(y_{i}\right)}-\frac{n-y_{i}}{\xi-y_{i}}>0

as a result of the classical inequality between the arithmetic mean and the harmonic mean.

It follows that
(17)

\operatorname{sg}g^{\prime}\left(y_{i}\right)=(-1)^{n-i}

and relations (16) and (17) demonstrate the property.
Note that the property remains true also for $y_{k}$ even if $y_{k}=x_{k+1}$ , provided, of course, that one has $x_{k+1}<x_{n}$ ,

So we have
LEMMA 2. - If the zeros $x_{k+1},x_{k+2},\ldots,x_{n}$ are all replaced by their arithmetic mean and if $x_{k+1}\left\langle x_{n}\right.$ , the zero $y_{k}$ and all zeros $y_{1},y_{2},\ldots,y_{k-1}$ which do not coincide with a $x_{i}$ become bigger.

By the transformation of $x$ in $-x11$ also results in the following property

Lemma 3. - If the zeros $x_{1},x_{2},\ldots,x_{k}$ are replaced by their arithmetic mean and $x_{1}<x_{k}$ , the zero $y_{k}$ and all zeros $y_{k+1},y_{k+2}\ldots,y_{n-1}$ which do not coincide with a $x_{i}$ become smaller.

As an application we can find Theorem 3 by simultaneously applying Lemmas 1, 2 and 3.
6. - Before going further we will establish some limitations relating to the polynomial

f(x)=(x-a)^{\lambda_{1}}(x-b)^{\lambda_{2}}(x-c)^{\lambda_{3}},a\leqq b\leqq c.

In this case a linear and homogeneous inequality between the zeros (1) and (2) is of the form

p_{1}a+p_{2}b+p_{3}c+q_{1}\alpha+q_{2}\beta\geqq 0.

Or $\alpha,\beta$ are respectively the zeros between $a$ And $b$ and between $b$ And $c$ of $f^{\prime}(x)$ (the zeros $y_{\lambda_{1}}$ And $y_{\lambda_{1}+\lambda_{2}}$ ).

We must necessarily have $p_{1}+p_{2}+p_{3}+q_{1}+q_{2}=0$ , as is easy to see. As a result of equality (3) and the homogeneity of the formula we can take any one of the coefficients equal to 0 and any one of the positive coefficients equal to 1 or any one of the negative coefficients equal to -1 ⁵ ). We can therefore take $q_{1}=0$ And $q_{2}=+1$ or -1 depending on what then $g_{2}>0$ Or $<0$ ,

This being so, we first have
THEOREM 4.-If the numbers $p_{1},p_{2},p_{3}$ check the conditions

p_{1}+p_{2}+p_{3}=-1,p_{1}\leqq 0,p_{3}\left(\lambda_{2}+\lambda_{3}\right)+\lambda_{2}\geqq 0,

we have inequality

p_{1}a+p_{2}b+\dot{p}_{3}c+\beta\geqq 0,

(19)

for any polynomial (18).
Indeed, according to inequality (6) we have

\beta\geq\frac{\lambda_{3}b+\lambda_{2}c}{\lambda_{2}+\lambda_{3}}

and we can write
$p_{1}a+p_{2}b+p_{3}c+\beta=\left(\beta-\frac{\lambda_{3}b+\lambda_{1}c}{\lambda_{2}+\lambda_{3}}\right)+\left(p_{3}+\frac{\lambda_{2}}{\lambda_{2}+\lambda_{3}}\right)(c-b)-p_{1}(b-a)$
ef Theorem 4 results from this.
We see moreover that in (19), the equality only takes place in one of the following cases

\begin{array}[]{r}a=b=c,\\ b=c_{1}p_{1}=0.\end{array}

⁰⁰footnotetext: 5. The trivial case

p_{1}=p_{2}=p_{3}=q_{1}=q_{2}=0

is obviously of no interest.

We also have
Theorem 5.-If the numbers $p_{1},p_{2},p_{3}$ check the conditions

\begin{gathered}p_{1}+p_{2}+p_{3}=-1,2p_{1}\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\leqq 0\right.\\ p_{3}\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right)+\lambda_{1}+\lambda_{2}\leqq 0\end{gathered}

we still have inequality (19) for any polynomial (18).
Indeed, according to lemma 3, we have

\beta\geqq\frac{2\left(\lambda_{1}+\lambda_{2}\right)c+\lambda_{3}(a+b)}{2\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right)}

and we can write

\begin{gathered}p_{1}a+p_{2}b+p_{3}c+\beta=\left[\beta-\frac{2\left(\lambda_{1}+\lambda_{2}\right)c+\lambda_{3}(a+b)}{2\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right)}\right]+\\ +\left(p_{3}+\frac{\lambda_{1}+\lambda_{2}}{\lambda_{1}+\lambda_{2}+\lambda_{3}}\right)(c-b)-\left(p_{1}+\frac{\lambda_{3}}{2\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right)}\right)(b-a)\end{gathered}

which proves Theorem 5.
In this case the equality in (19) only holds in one of the following cases

\begin{gathered}a=b=c,\\ a=b,p_{3}\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right)+\lambda_{1}+\lambda_{2}=0.\end{gathered}

Finally, let us also demonstrate
Theorem 6. - For inequality
(20)

p_{1}a+p_{2}b+p_{3}c-\beta\geqq 0

is verified whatever the polynomial (18), it is necessary and sufficient that we have

p_{1}+p_{2}+p_{3}=1,p_{1}\leqq 0,p_{3}\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right)-\left(\lambda_{1}+\lambda_{2}\right)\geqq 0

We first see that the conditions are necessary by doing $b=c($ SO $b=c=\beta)$ And $a=b\left(\right.$ and then $\left.\beta=\frac{\lambda_{3}b+\left(\lambda_{1}+\lambda_{2}\right)c}{\lambda_{1}+\lambda_{2}+\lambda_{3}}\right)$ .

To see that the conditions are also sufficient, note that, according to inequality (6), we have

\beta\leqq\frac{\lambda_{3}b+\left(\lambda_{1}+\lambda_{2}\right)c}{\lambda_{1}+\lambda_{2}+\lambda_{3}}.

We can write

\begin{gathered}p_{1}a+p_{2}b+p_{3}c-\beta=\left(\frac{\lambda_{3}b+\left(\lambda_{1}+\lambda_{2}\right)c}{\lambda_{1}+\lambda_{2}+\lambda_{3}}-\beta\right)+\left(p_{3}-\frac{\lambda_{1}+\lambda_{2}}{\lambda_{1}+\lambda_{2}+\lambda_{3}}\right)(c-b)-\\ -p_{1}(b-a)\end{gathered}

from which it follows that the conditions are also sufficient.
7. - The reader has certainly noticed the essential difference between the theorems $\mathbf{4},\mathbf{5}$ on the one hand and the theorem $\mathbf{6}$ on the other hand. While the first two express only sufficient conditions, the third gives conditions that are both necessary and sufficient.

So let us return to inequality (19). We find, as for inequality (20), the necessary conditions

p_{1}\leqq 0,\quad p_{3}\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right)+\lambda_{1}+\lambda_{2}\geqq 0.

(21)

To completely solve the problem let us note that we can take

a=-1,\quad b=0,\quad c=h\geqq 0

Let us designate by $E(h)$ the first member of (19) multiplied by the positive number $\frac{2\left(\lambda_{1}+\lambda_{1}+\lambda_{3}\right)}{\lambda_{1}+\lambda_{2}}$ . If we notice that $\beta$ is the non-negative root of the equation

\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right)x^{2}-\left[\left(\lambda_{1}+\lambda_{2}\right)h-\lambda_{2}-\lambda_{3}\right]x-\lambda_{2}h=0

and if we do the calculations we find

E(h)=V\overline{(h+A)^{2}+B^{2}-Ch-D}

where
(22) $\left\{\begin{array}[]{l}A=\frac{\left(\lambda_{1}+\lambda_{2}\right)\left(\lambda_{2}+\lambda_{3}\right)-2\lambda_{1}\lambda_{3}}{\left(\lambda_{1}+\lambda_{2}\right)^{2}},\quad B=\frac{2\sqrt{\lambda_{1}\lambda_{2}\lambda_{3}\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right)}}{\left(\lambda_{1}+\lambda_{2}\right)^{2}}\\ C=-\frac{2p_{2}\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right)}{\lambda_{1}+\lambda_{2}}-1,\quad D=\frac{2p_{1}\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right)}{\lambda_{1}+\lambda_{2}}+\frac{\lambda_{2}+\lambda_{3}}{\lambda_{1}+\lambda_{2}}.\end{array}\right.$

We must find the necessary and sufficient conditions so that we have

E(h)\geqq 0

(23)

regardless of $h\geqq 0$ .

With the notations (22) the necessary options (21) become

\sqrt{A^{2}+B^{2}}-D\geqslant 0,\quad 1\geqq C

and express that the inequality (23) is indeed verified for $h=0$ and for $h$ infinitely large, if

C\leqslant\frac{A}{\sqrt{A^{2}+B^{2}}}

we are in the conditions of thegreme 4.
It remains to examine the case
(24) $1\geqq C\geqq\frac{A}{\sqrt{A^{2}+B^{2}}}\left(\right.$ Or $\left.-\frac{\lambda_{2}}{\lambda_{2}+\lambda_{3}}>p_{3}\geqq-\frac{\lambda_{1}+\lambda_{2}}{\lambda_{1}+\lambda_{2}+\lambda_{3}}\right)$ .

We have

\frac{d^{3}E(h)}{dh^{2}}=\frac{B^{3}}{\left(V(h+A)^{2}+B^{2}\right)^{8}}\geqslant 0

which shows us that the derivative $\frac{dE(h)}{dh}$ is always increasing.

\frac{dE(h)}{dh}=\frac{h+A}{\sqrt{(h+A)^{2}+B^{2}}}-C

it follows that this derivative only becomes zero for

h=h_{1}=-A+\frac{CB}{\sqrt{1-C^{2}}},

As a result of inequalities (24), $h_{1}$ is indeed a positive number.

For inequality (23) to be satisfied it is necessary and sufficient that $E\left(h_{1}\right)\geqq 0$ . We have

E\left(h_{1}\right)=AC+B\sqrt{1-C^{2}-D}

donc la condition cherchée est

AC+B\sqrt{1-C^{2}-D}\geqq 0.

Il est facile de voir que le résultat subsiste aussi pour $C=1$ , - lorsque $h_{1}$ n’existe pas.

Finalement nous pouvons énoncer la propriété suivante.
THEORÈME. 7. - Pour que l’inégalité (19) ait lieg pour tout
polynome (18) il faut et il suffit que l’on ait $p_{1}+p_{2}+p_{3}=-1$ et, ou bien

p_{1}\leqq 0,p_{3}\left(\lambda_{2}+\lambda_{3}\right)+\lambda_{2}\geqq 0.

qu bien

	$\displaystyle AC+BV\overline{1-C^{2}}-D\geqq 0,p_{3}\left(\lambda_{2}+\lambda_{3}\right)+\lambda_{2}<0,$		(25)
	$\displaystyle p_{3}\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right)+\lambda_{1}+\lambda_{2}\geqq C.$

$A,B,C,D$ étant définis par les formules (22).
Dans le szcond cas l’égalité n’a lieu que si
$AC+B\sqrt{1-C^{2}}-D=0,\sqrt{1-C^{2}}(c-b)=(b-a)\left(CB-A\sqrt{1-C^{2}}\right)$ .
8. - On voit que les théorèmes 4 et 5 ne nous donnent pas des propriétés plus complètes que celles exprimées par les théorèmes 1 et 3 . Le théorème 6 précise un peu la propriété correspondante exprimée par le théorème 1.

En ce qui concerne la valeur du théorème 7, elle est toute autre. Ce théorème nous donne effectivement des inégalités qui ne sont pas toujours comprises dans les théorèmes 1 et 3. Soit, en effet, l’inégalité (19) où $p_{1},p_{3}$ vérifient les conditions (25). Par suite de l’inégalité

p_{3}+\frac{\lambda_{2}}{\lambda_{2}+\lambda_{3}}<0,

l’inégalité (19) considérée n’est pas une conséquence du théorème 1. Pour voir que notre inégalité peut ne pas être une conséquence du théorème 3, il suffit de montrer que les inégalités (25) sont compatibles avec l’inégalité

p_{1}+\frac{\lambda_{3}}{2\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right)}>0.

Pour celail faul et il suffit de montrer que les inégalités

\frac{\lambda_{2}}{\lambda_{2}+\lambda_{3}}<AC+B\sqrt{1-C^{2}},\quad 1\geqq C>\frac{A}{\sqrt{A^{2}+B^{2}}}

sont compatibles. Cette condition de compatibilité est

\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}>\sqrt{A^{2}+B^{2}}=\frac{\lambda_{2}+\lambda_{3}}{\lambda_{1}+\lambda_{2}}

et elle est bien vérifiée
9. - Nous n’avons pas l’intention d’étudier toutes les conséquences des résultafs précédents dans le cas d’un polynome quelconque $f(x)$ . Nous allons seulement signaler quelques inégalités qui nous seront utiles dans le § suivant,

Soit donc $f(x)$ un polynome quelconque à zéros tous réels et considérons une inégalité de la forme
où

r_{1}\sum_{v=1}^{j}x_{v}+r_{2}\sum_{v-j+1}^{n}x_{v}-s\left(y_{1}+y_{2}+\cdots+y_{j-1}\right)-s^{\prime}y_{j}\geqq 0

(26)

ir_{1}+(n-j)r_{2}=(j-1)s+s^{\prime},s>s^{\prime}>0

(27)

En vertu du lemme 2, si l’inégalité (26) est vérifiée lorsqu’on remplace $x_{j+1},x_{j+2},\ldots,x_{n}$ par leur moyenne arithmétique, elle sera vérifiée pour tout polynome $f(x)$ . Il en résulte qu’il suffit de démontrer l’inégalité (26) pour $x_{j+1}=x_{j+2}=\cdots=x_{n}$ .

Dans ce dernier cas elle s’écrit :

r_{1}\sum_{v_{-1}}^{j}x_{v}+\left(n-jr_{2}x_{j+1}-s\left(y_{1}+y_{2}+\cdots+y_{j-1}\right)-s^{\prime}y_{j}\geqq 0.\right.

Mais, l’égalité (3) nous donne

y_{1}+y_{2}+\cdots+y_{j-1}=\frac{n-1}{n}\sum_{v=1}^{j}x_{v}+\frac{j}{n}x_{j+1}-y_{j}

et notre inégalité devient
(28) $\left[nr_{1}-s(n-1)\right]\sum_{v=1}^{j}x_{j}+\left[n(n-j)r_{2}-sj\right]x_{j+1}+n\left(s-s^{\prime}\right)y_{j}\geqq 0$ .

En vertu du lemme 3, si l’inégalité (28) est vérifiée lorsque, de plus, on remplace $x_{1},x_{2},\ldots,x_{j}$ par leur moyenne arithmétique, elle sera toujours vérifiée. Il suffit donc de démontrer l’inégalité (28) si de plus, $x_{1}=x_{2}=\cdots=x_{j}$ . L’inégalité devient alors
(29) $\left[nr_{1}-s(n-1)\right]jx_{j}+\left[n(n-j)r_{2}-sj\right]x_{j+1}+n\left(s-s^{\prime}\right)y_{j}\geqq 0$ .

Mais, nous avos maintenant,

y_{j}=\frac{(n-j)x_{j}+jx_{j+1}}{n}

et l’inégalité ( 29 ) devient

\left[n(n-j)r_{2}-js^{\prime}\right]\left(x_{i+1}-r_{j}\right)\geqq 0

Nous en déduisons donc le
Théorème 8. - Pour que l’inégalité (26), sous les conditions (27), ait lieu pour tout polynome $f(x)$ , à zéros tous réels, il faut et il suffit que l’on ait

n(n-j)r_{2}-js^{\prime}\geqq 0

Dans le cas $j=1$ , il faut prendre $s=0$ dans les formules (26) et (27) et $s^{\prime}>0$ . Sous les conditions données, l’égalité dans (26) n’a lieu que pour $x_{1}=x_{2}=\cdots=x_{i},x_{i+1}==x_{j+2}=\cdots=x_{n}$ lorsque $n(n-j)r_{2}-is^{\prime}=0$ et seulement si $x_{1}=x_{2}=\cdots=x_{n}$ lorsque $n(n-i)r_{2}-js^{\prime}>0$ .
10. - Nous allons étudier maintenant des inégalités un peu plus générales. Considérons l’inégalité
(30) $r_{1}\sum_{v=1}^{j-1}x_{v}+r_{2}x_{j}+r_{3}\sum_{v=j+1}^{n}x_{v}-s\left(1_{1}+y_{2}+\cdots+y_{j}\right)+s^{\prime}y_{j-1}+s^{\prime\prime}y_{j}\geqq 0$ , où

	$\displaystyle(j-1)r_{1}+r_{2}+(n-j)r_{3}=js-s^{\prime}-s^{\prime\prime},s\geqq s^{\prime\prime}>s^{\prime}\geqq 0$		(31)
	$\displaystyle 1<i\leqq n-1$

Comme plus haut, nous voyons que, en vertu du lemme 2, il suffit de dénontrer l’inégalité si $x_{j+1}=x_{j+2}=\cdots=x_{n}$ . The inequality then becomes

\left(r_{1}-s\frac{n-1}{n}\right)_{v=1}^{i-1}x_{v}+\left(r_{2}-s\frac{n-1}{n}\right)x_{j}+\left[(n-j)r_{3}-s\frac{j}{n}\right]x_{j+1}+s^{\prime}y_{j-1}+s^{\prime\prime}y_{j}\geq 0

Similarly, by virtue of Lemma 3, it is sufficient to prove this inequality if, in addition, we have $x_{1}=x_{2}=\cdots=x_{j-1}$ . In this last case equality (3) still gives us

y_{j-1}=\frac{n-j+1}{n}x_{j-1}+\frac{n-1}{n}x_{j}+\frac{j}{n}x_{j+1}-y_{j}

and our inequality becomes
(32) where

p_{1}x_{j-1}+p_{2}x_{j}+p_{3}x_{j+1}+y_{j}\geqq 0,

\text{ (33) }\left\{\begin{array}[]{l}p_{1}=\frac{1}{n\left(s^{\prime\prime}-s^{\prime}\right)}\left[n(j-1)r_{1}-(n-1)(j-1)s+(n-j+1)s^{\prime}\right]\\ p_{2}=\frac{1}{n\left(s^{\prime\prime}-s^{\prime}\right)}\left[nr_{2}+(n-1)\left(s^{\prime}-s\right)\right]\\ p_{3}=\frac{1}{n\left(s^{\prime\prime}-s^{\prime}\right)}\left[n(n-j)r_{3}+j\left(s^{\prime}-s\right)\right]\end{array}\right.

Inequality (32) is of the form (19) corresponding to the polynomial (18) where
$\lambda_{1}=j-1,\quad\lambda_{2}=1,\quad\lambda_{3}=n-j,\quad a=x_{j-1},\quad b=x_{j},\quad c=x_{j+1}$ .
Finally we therefore deduce
THEOREM 9. - For inequality (30), under conditions (31), to hold for any polynomial $f(x)$ with all real zeros, it is necessary and sufficient that we have either
(34)
or

p_{1}=0,(n-j+1)p_{3}+1\geqq 0.

(35) $-AC+B\sqrt{1-C^{2}}-D=0,\quad(n-j+1)p_{3}-1<0,\quad np_{3}+j\geqq 0$ , Or

\left\{\begin{array}[]{l}A=\frac{n-(j-1)(n-j)}{j^{2}},\quad B=\frac{2\sqrt{n(j-1)(n-j)}}{j^{2}}\\ C=-\frac{2np_{3}+1}{i},\quad D=\frac{2np_{1}+n-j+1}{j}\end{array}\right.

$p_{1},p_{3}$ given by formulas (33).
We can easily obtain the equality conditions in (30), but there is no need to insist on this point here. Note also that Theorem 5 shows us that the conditions

2np_{1}+n-j\leqq 0,\quad np_{3}+j\geqq 0,

are sufficient for inequality (30).

§ 3.

11.
- —
  
  Consider two sequences of real numbers

		$\displaystyle\alpha_{1},\alpha_{2},\ldots,\alpha_{m},$
		$\displaystyle\alpha_{1}^{\prime},\alpha_{2}^{\prime},\ldots,\alpha_{m}^{\prime}.$

We will say that the sequence (38) is an average sequence of the sequence (37) if we can find $m^{2}$ non-negative numbers $p_{i,j},i,j=1,2,\ldots,m$ such that we have

\sum_{j=1}^{m}\hat{p}_{i,j}=\sum_{j=1}^{m}p_{j,i}=1,\alpha_{i}^{\prime}=\sum_{j=1}^{m}p_{i,j}\alpha_{j},i=1,2,\ldots,m.

This condition does not depend on the order of the terms of the sequences (37), (38). We can therefore assume that

\alpha_{1}\leqq\alpha_{2}\leqq\cdots\leqq\alpha_{m},\alpha_{1}^{\prime}\leqq\alpha_{2}^{\prime}\leqq\cdots\leqq\alpha_{m}^{\prime}

(39)

Messrs. G. H, HARDY, JE LITfLEWOOD and G. POLYA have demonstrated ⁶ ) the following property, expressed by the

LEMMA 4.-For the sequence (38) to be an average sequence of the sequence (37), it is necessary and sufficient that we have

	$\displaystyle\alpha_{1}^{\prime}+\alpha_{2}^{\prime}+\cdots+\alpha_{m}^{\prime}\equiv\alpha_{1}+\alpha_{2}+\cdots+\alpha_{m}$		(40)
	$\displaystyle\alpha_{1}^{\prime}+\alpha_{2}^{\prime}+\cdots+\alpha_{j}^{\prime}\geqq\alpha_{1}+\alpha_{2}+\cdots+\alpha_{j};j=1,2,\ldots,m-1,$

assuming that the two lines are ordered in the manner (39). It is easy to see that if, instead of (39), we had

\alpha_{1}\geqq\alpha_{2}\geqq\cdots\geqq\alpha_{m},\alpha_{1}^{\prime}\geqq\alpha_{2}^{\prime}\geqq\cdots\geqq\alpha_{m}^{\prime}

(

\prime

)

we would have, instead of (41), the inequalities
(41') $\alpha_{1}^{\prime}+\alpha_{2}^{\prime}+\cdots+\alpha_{j}^{\prime}\leq\alpha_{1}+\alpha_{2}+\cdots+\alpha_{j};j=1,2,\ldots,m-1$ .
12 - Let us take the sequence (1) and its derived sequence (2). Let us define the sequences (37), (38), by taking $m=n(n-1)$ , in the following manner

\begin{gathered}\alpha_{in-n-i+2}=\alpha_{in-n-i+3}=\cdots=\alpha_{in-i}=x_{i},i=1,2,\ldots,n;\\ \alpha_{jn-n+1}^{\prime}=\alpha_{jn-n+2}^{\prime}=\cdots=\alpha_{jn}^{\prime}=y_{y},j=1,2,\ldots,n-1.\end{gathered}

6.

GH HARDY, JE LittlafwOOD, G. POLYA „Inequalities" Cambridge Univ. Press, 1934, Chap. II.

Equality (40) is then verified and returns to (3). We have (39) and inequalities (41) are written ⁷ )

\text{ (42) }\left\{\begin{array}[]{c}n\left(y_{1}+y_{2}+\cdots+y_{j-1}\right)+ry_{j}\geqq(n-1)\left(x_{1}+x_{2}+\cdots+x_{j-1}\right)+(r+j-1)x_{j},\\ r=1,2,\ldots,n-i,j=1,2,\ldots,n-1,\\ n\left(y_{1}+y_{2}+\cdots+y_{j-1}\right)+ry_{j}\geqq(n-1)\left(x_{1}+x_{2}+\cdots+x_{j}\right)+(r+j-n)x_{j+1},\\ r=n-j+1,n-j+2,\ldots,n,j=1,2,\ldots,n-1,\end{array}\right.

\text{ (sauf }r=n,j=n-1\text{ ), }

Consider, in particular, the inequalities
(43)

\begin{gathered}n\left(y_{1}+y_{2}+\cdots+y_{j}\right)\geqq(n-1)\left(x_{1}+x_{2}+\cdots+x_{j}\right)+jx_{j+1},\\ j=1,2,\ldots,n-2.\end{gathered}

All other inequalities (42) are consequences of these and of inequalities (4).

But, the inequalities (43) are none other than the inequalities (12) for $i=1$ . We can therefore state the following property:

THEOREM 10. - The derived sequence (2), where each term is repeated $n$ times, is an average sequence of the primitive sequence (1) in which each term is repeated $n-1$ times.

From this property it follows that we can find $n(n-1)$ non-negative numbers $q_{r,s},r=1,2,\ldots,n-1,s=1,2,\ldots,n$ so that we have

\begin{gathered}y_{j}=\sum_{r=1}^{n}q_{r,j}x_{r},\sum_{r=1}^{n}q_{r,j}=1,\sum_{s=1}^{n-1}q_{i,s}=\frac{n-1}{n},\\ i=1,2,\ldots,n,j=1,2,\ldots,n-1.\end{gathered}

In a previous work ⁸ ) we established this property by noticing that

⁰⁰footnotetext: 7. When

j=1

the sums such as

x_{1}+x_{2}+\cdots+x_{j-1},y_{1}+y_{2}++\cdots+y_{j-1}

are replaced by zero. 8i Tiberiu Popoviciu, „Notes on convex functions of higher order" (III) ", Mathematica, 16, 74-86 (1940).

y_{j}=\frac{\sum_{r=1}^{n}\frac{x_{r}}{\left(y_{j}-x_{r}\right)^{2}}}{\sum_{r=1}^{n}\frac{1}{\left(y_{j}-x_{r}\right)^{2}}},j=1,2,\ldots,n-1

and demonstrating that

\sum_{s=1}^{n-1}\frac{\frac{1}{\left.\sum_{t=1}^{n}-x_{i}\right)^{2}}}{\frac{1}{\left(y_{s}-x_{r}\right)^{2}}}=\frac{n-1}{n},\quad i=1,2,\ldots,n.

This is how we first obtained the inequalities $\left.(43)^{9}\right)$ .
13. - Let us again consider the sequence (i) and its derived sequence
(2). Let us now set
(44) $\left\{\begin{array}[]{l}x_{i}^{\prime}=\frac{x_{1}+x_{2}+\cdots+x_{i-1}+x_{i+1}+\cdots+x_{n}}{n-1},\quad i=1,2,\ldots,n.\\ y_{j}^{\prime}=\frac{y_{1}+y_{2}+\cdots+y_{i-1}+y_{j+1}+\cdots+y_{n-1}}{n-2},\quad j=1,2,\ldots,n-1,\end{array}\right.$
assuming, of course, $n>2$ .
In this way we obtain the sequences

	$\displaystyle x_{1}^{\prime}\geqq x_{2}^{\prime}\geqq\cdots\geq x_{n}^{\prime}$		(45)
	$\displaystyle y_{1}^{\prime}\geqq y_{2}^{\prime}\geqq\cdots\geqq y_{n-1}^{\prime}$		(46)

and equality

\frac{1}{n}\sum_{i=1}^{n}x_{i}^{\prime}=\frac{1}{n-1}\sum_{i=1}^{n-1}y_{j}^{\prime}

(47)

Theorem 2 shows us that
(48)

y_{1}^{\prime}\geqq x_{1}^{\prime},x_{n}^{\prime}\geqq y_{n-1}^{\prime}.

Let us now define the sequences (37), (38) in the following way, again taking $m=n(n-1)$ ,

⁰⁰footnotetext: 9. See loc. cit. 8).

		$\displaystyle\alpha_{in-n+1}=\alpha_{in-n+2}=\cdots=\alpha_{in}=y_{j}^{\prime},\quad j=2,\ldots,n-1$
		$\displaystyle\alpha_{in-n-i+2}^{\prime}=\alpha_{in-n-i+3}^{\prime}=\cdots=\alpha_{in-i}^{\prime}=x_{i}^{\prime},\quad i=2,\ldots,n$

and let us check whether the conditions of Lemma 4 are verified.
Equality (40) is satisfied by following (47).
This time we have ( $39^{\prime}$ ) and so we need to look at the inequalities ( $41^{\prime}$ ). These inequalities become ¹⁰ ).
(49)

\left\{\begin{array}[]{c}n\left(y_{1}^{\prime}+y_{2}^{\prime}+\cdots+y_{j-1}^{\prime}\right)+ry_{j}^{\prime}\geqq(n-1)\left(x_{1}^{\prime}+x_{2}^{\prime}+\cdots+x_{j-1}^{\prime}\right)+(r+j-1)x_{j}^{\prime}\\ r=1,2,\ldots,n-j,\quad j=1,2,\ldots,n-1\\ n\left(y_{1}^{\prime}+y_{2}^{\prime}+\cdots+y_{j-1}^{\prime}\right)+ry_{j}^{\prime}\geqq(n-1)\left(x_{1}^{\prime}+x_{2}^{\prime}+\cdots+x_{j}^{\prime}\right)+(r+j-n)x_{j+}^{\prime}\\ r=n-j+1,\quad n-j+2,\ldots,n,\quad j=1,2,\ldots,n-1\\ (\text{ sauf }r=n,\quad j=n-1)\end{array}\right.

From these inequalities let us choose the following
(50) $n\left(y_{1}^{\prime}+y_{2}^{\prime}+\cdots+y_{j-1}^{\prime}\right)+y_{j}^{\prime}\geqq(n-1)\left(x_{1}^{\prime}+x_{2}^{\prime}+\cdots+x_{j-1}^{\prime}\right)+jx_{j}^{\prime}$

j=2,3,\ldots,n-2,

(51) $n\left(y_{1}^{\prime}+y_{2}^{\prime}+\cdots+y_{j-1}^{\prime}\right)+(n-j)y_{j}^{\prime}\geqq(n-1)\left(x_{1}^{\prime}+x_{2}^{\prime}+\cdots+x_{j}^{\prime}\right)$

j=2,3,\ldots,n-2,

(52) $n\left(y_{1}^{\prime}+y_{2}^{\prime}+\cdots+y_{j-1}^{\prime}\right)+(n-j+1)y_{j}^{\prime}\geqq(n-1)\left(x_{1}^{\prime}+x_{2}^{\prime}+\cdots+x_{j}^{\prime}\right)+x_{j+1}^{\prime}$

j=2,3,\ldots,n-2,

(53) $n\left(y_{1}^{\prime}+y_{2}^{\prime}+\cdots+y_{j}^{\prime}\geqq(n-1)\left(x_{1}^{\prime}+x_{2}^{\prime}+\cdots+x_{j}^{\prime}\right)+jx_{j+1}^{\prime}\right.$

j=2,3,\ldots,n-2.

All other inequalities (49) are consequences of these inequalities, of inequalities (45), (46), (48) and of equality (47). Let us note, in passing, that for $n=3$ inequalities (49) are demonstrated. It remains to demonstrate inequalities (50), (51), (52), and (53) for $n>3$ . First we will express these inequalities using the zeros (1) and (2), taking into account (44). Doing the calculations we find
10) For $j=1$ the same covenants as above. See 7).
(54)

[n(j-1)+1]\sum_{i=1}^{n}x_{i}+n(n-1)(n-2)\left(x_{1}+x_{2}+\cdots+x_{j-1}\right)+n(n-2)jx_{i}-

j=2,3,\ldots,n-2

(55) $j\sum_{i=1}^{n}x_{j}+n(n-2)\left(x_{1}+x_{2}+\cdots+x_{j}\right)-n^{2}\left(y_{1}+y_{2}+\cdots+y_{j}\right)+jny_{j}\geqq 0$

j=2,3,\ldots,n-2

[(j-1)(n-1)+1]\sum_{i=1}^{n}x_{i}+n(n-1)(n-2)\left(x_{1}+x_{2}+\cdots+x_{j-1}\right)+

	$\displaystyle+n(n-2)x_{j}-n^{2}(n-1)\left(y_{1}+y_{2}+\cdots+y_{j}\right)+n(n-1)(j-2)y_{j-1}+$		(56)
	$\displaystyle+n^{2}(n-1)y_{j}\geqq 0$
	$\displaystyle j=3,4,\ldots,n-1$

(57)

\begin{gathered}(j-1)\sum_{i=1}^{n}x_{i}+(n-1)(n-2)\left(x_{1}+x_{2}+\cdots+x_{j-1}\right)+\\ +(j-1+1)(n-2)x_{j}-n(n-1)\left(y_{1}+y_{2}+\cdots+y_{j}\right)+n(n-1)y_{j}\geqq 0,\\ i=3,4,\ldots,n-1.\end{gathered}

Before doing the calculations in (52) and (53) we changed $j$ in $j-1$ .

It remains for us to demonstrate inequalities (54), (55), (56), (57).
14. - Let us first deal with inequality (55), which is the simplest.

Inequality (55) is of the form (26), where we have

r_{1}=j+n(n-2),r_{2}=j,s=n^{2},s^{\prime}=n(n-j)

and conditions (27) are indeed verified.
We have

n(n-j)r_{2}-js^{\prime}=0

and we can therefore apply Theorem 8.
Inequality (55) is therefore demonstrated.
15. - Inequalities (54), (56), (57) are of the form (30).
I. Inequality (54). We have
$r_{1}=n(j-1)+1+n(n-1)(n-2),r_{2}=n(j-1)+1+n(n-2)j=(n-1)(nj-1)$ ,

r_{3}=n(j-1)+1,s=n^{2}(n-1),s^{\prime}=0,s^{\prime\prime}=n(n-1)^{2}

Conditions (31) are indeed verified.
For numbers (33) we have

\begin{gathered}p_{1}=-\frac{(j-1)\left(n^{2}-nj-1\right)}{n(n-1)^{2}},p_{2}=-\frac{n(n-1)+1-nj}{n(n-1)}\\ p_{3}=-\frac{nj^{2}-(2n-1)}{n(n-1)^{2}}\end{gathered}

We have

(n-j+1)p_{3}+1=\frac{(j-1)\left[nj^{2}-\left(n^{2}+2n-1\right)j+2n(n-1)\right]}{n(n-1)^{2}}

and it is easy to verify that for $2\leqq j\leqq n-1$ This number is negative. The second condition (34) is therefore not verified.

We have

np_{3}+j=\frac{n(j-1)}{(n-1)}(n-j-1)>0,2\leqq j\leqq n-2

and it remains for us to examine the first inequality (35).
Taking into account (36), we find

AC-D=-\frac{4n(j-1)(n-j)(n-j-1)}{(n-1)^{2}j^{\prime}}

$\sqrt{1-C^{2}}=\frac{2}{j}\sqrt{-np_{3}\left(np_{3}+j\right)}=\frac{2}{(n-1)^{2}}-\sqrt{n(j-1)(n-j-1)\left[nj^{2}-(2n-1)j+n(n-1)\right]}$
and the first inequality (35) becomes, after simplifications,
or

\sqrt{nj^{2}-(2n-1)j+n(n-1)}\geqq\sqrt{(n-j)(n-i-1)}

(n-1)j^{2}\geqq 0

which is obviously verified.
We can therefore apply Theorem 9 and inequality (54) is demonstrated.
II. Inequality (56). We proceed exactly as for (54). We have

\begin{gathered}r_{1}=(j-1)(n-1)+1+n(n-1)(n-2),r_{2}=(j-1)(n-1)+1+n(n-2)=(n-1)(n+j-2)\\ r_{3}=(j-1)(n-1)+1,s=n^{2}(n-1),s^{\prime}=n(n-1)(j-2),s^{\prime\prime}=n^{2}(n-1)\end{gathered}

We deduce

\begin{gathered}p_{1}=-\frac{n-j}{(n-1)(n-j+2)},p_{2}=-\frac{n-j}{n-j+2},p_{3}=-\frac{n+j-2}{(n-1)(n-j+2)}\\ (n-j+1)p_{3}+1=-\frac{(n-j)(j-2)}{(n-1)(n-j+2)}<0,3\leqq j\leqq n-1\end{gathered}

and the second inequality (34) is not verified.
We also have

\begin{gathered}np_{3}+j=\frac{(n-j)[(n-1)j-(n-2)]}{(n-1)(n-j+2)}>0,3\leqq j\leqq n-1,\\ AC-D=-\frac{4n(n-2)(j-1)(n-j)}{(n-1)(n-j+2)j},\\ \sqrt{1-C^{2}}=\frac{2}{(n-1)(n-j+2)j}\sqrt{n(n-j)(n+j-2)[(n-1)j-(n-2)]}\end{gathered}

and the first inequality (35) becomes, after simplifications,

\sqrt{(n+j-2)[(n-1)j-(n-2)]}\geqq(n-2)\sqrt{j-1}

(n-1)j^{2}\geqq 0

By Theorem 9, inequality (56) is therefore proven.
III. Inequality (57). We have

\begin{gathered}r_{1}=(j-1)+(n-1)(n-2),r_{2}=(j-1)+(j-1)(n-2)=(n-1)(j-1)\\ r_{3}=j-1,s=n(n-1),s^{\prime}=0,s^{\prime\prime}=n(n-1)\end{gathered}

and we deduce

\begin{gathered}p_{1}=-\frac{(j-1)(n-j)}{n(n-1)},p_{2}=-\frac{n-j}{n},p_{3}=-\frac{j^{2}-2j+n}{n(n-1)}\\ (n-i+1)p_{3}+1=-\frac{(j-1)(j-2)(n-j)}{n(n-1)}<0,3\leqq j\leqq n-1\end{gathered}

The second inequality (34) is therefore not verified.
We also have

\begin{gathered}np_{3}+j=\frac{(j-1)(n-j)}{n-1}>0,3\leqq j\leqq n-1\\ AC-D=-\frac{4(j-1)(n-j)^{2}}{(n-1)j^{3}}\\ \sqrt{1-C^{2}}=\frac{2}{(n-1)j}\sqrt{(j-1)(n-j)\left(j^{2}-2j+n\right)}\end{gathered}

and the first inequality (35) becomes
or

\sqrt{n\left(j^{2}-2j+n\right)}\geqq n-j

(n-1)j^{2}\geqq 0

which is obviously verified.
Inequality (57) is therefore also demonstrated.
16. - Finally we obtained the following property, analogous to that expressed by Theorem 10.

Theorem 11. - The sequence (45) where each term is repeated $n-1$ times is an average sequence of the sequence (46) in which each term is repeated $n$ times ( $n>2$ ).

As a result, we can find $n(n-1)$ non-negative numbers $q_{r,s}^{\prime},r=1,2,\ldots,n,s=1,2,\ldots,n-1$ , so that we have

\begin{gathered}x_{i}^{\prime}=\sum_{r=1}^{n-1}q_{i,r}^{\prime}y_{r}^{\prime},\sum_{z=1}^{n-1}q_{i,r}^{\prime}=1,\sum_{s=1}^{n}q_{s,j}^{\prime}=-\frac{n}{n-1}\\ i=1,2,\ldots,n,j=1,2,\ldots,n-1\end{gathered}

$\S 4$ .

17.
- —
  
  Messrs. GH Hardy, JL Littlewood and G. Polya introduced the notion of average sequence by studying certain inequalities verified by convex functions ¹¹ ). This property, in a slightly more general form ¹² ), is stated as follows.

11.

GH Hardy, JE Littlewood, G. Polya „Some simple inequalities satisfied by convex functions ^∗ Messenger of Math., 58, 145-152 (1928). 12) See loc. cit. 8).

Lemma 5.-For the inequality

\sum_{i=1}^{r}A_{i}\varphi\left(x_{i}\right)\geqq\sum_{l=1}^{s}B_{i}\varphi\left(y_{i}\right),(r+s\geqq 3),

(58)

A_{i}>0,B_{i}>0,A_{1}-A_{2}+\cdots+A_{r}=B_{1}+B_{2}+\cdots+B_{s}=1,

$x_{1}<x_{2}<\cdots<x_{r},y_{1}<y_{2}<\cdots<y_{s}$ , be verified for any function $\varphi(x)$ non-concave (of order 1) in an interval containing the points $x_{i},y_{i}$ , it is necessary and sufficient that we can find rs non-negative numbers $p_{i,j}$ so that we have

\begin{gathered}\sum_{v=1}^{r}p_{i,v}=1,\sum_{v=1}^{r}B_{v}p_{v,j}=A_{j},y_{i}=\sum_{v=1}^{r}p_{i,v}x_{v},\\ i=1,2,\ldots,s,j=1,2,\ldots,r.\end{gathered}

If $\varphi(x)$ is convex, in (58) the sign $>$ is valid.
From this property we deduce
THEOREM 12. - If $\varphi(x)$ If a non-concave function in an interval containing (1) and if (2) is the derivative sequence of (1), we have the inequality

\frac{1}{n}\sum_{i=1}^{n}\varphi\left(x_{i}\right)\geqslant\frac{1}{n-1}\sum_{j=1}^{n=1}\varphi\left(y_{j}\right).

If the function $\varphi(x)$ is convex, equality only holds st $x_{1}=x_{2}=\cdots=x_{n}$ .

This property is due to MK TodA ¹³ ).
We pointed out this inequality some time ago, and it was actually demonstrated by MH E BRAY ¹⁴ ), for $\varphi(x)=xp,p$ positive integer. For $p$ any inequality has been studied by MS KaKeya ¹⁵ ).
18. - Lemma 5 can also be applied to the sequences (45) and (46). In this way we obtain a new property which is given by the
13) Kiyoshi toda "On certain functional inequalities" Journal of the Hiroshima Univ., (A), 4, 27-40 (1934).
14) Hubert E. Bray "On the zeros of a polynomial and of its derivative" Amer. Journal of Math, 55, 864-872 (1931).
15) Soichi Kakeya. On an inequality between the roots of an equation and its derivative" Proc. Phys. - Math. Soc. Japan, (3), 15, 149-154 (1933).

THEOREM 13. - If the sequences (45) and (46) are obtained from (1) and (2) by the formulas (44) and if $\varphi(x)$ is a nonconcave function in an interval containing the points (45) and (46), we have the inequality

\frac{1}{n-1}\sum_{j=1}^{n-1}\varphi\left(y_{j}^{\prime}\right)\geqq\frac{1}{n}\sum_{i=1}^{n}\varphi\left(x_{i}^{\prime}\right),\quad(n>2)

(59)

If the function is convex, equality only holds if $x_{1}=x_{2}==\cdots=x_{n}$ .

Indeed, from the study of Lemma 5 it results that equality cannot take place, if $\varphi(x)$ is convex, that if $y_{1}^{\prime}=y_{2}^{\prime}==\cdots=y_{n-1}^{\prime}$ , which requires $x_{1}=x_{2}=\cdots=x_{n}$ .

In particular, we have the inequality

\frac{y_{1}^{\prime p}+y_{2}^{\prime p}+\cdots+y_{n-1}^{\prime p}}{n-1}\geqq\frac{x_{1}^{\prime p}+x_{2}^{\prime p}+\cdots+x_{n}^{\prime p}}{n}

(60)

if $p>1$ , equality taking place only if $x_{1}=x_{2}=\ldots=x_{n}$ ; assuming, of course, that all zeros (1), and therefore also the numbers (45), (46) are non-negative. This results from the fact that the function $x^{p}$ is convex if $p>1,x\geqq 0$ . The function $x^{p}$ being concave if $0<p<1,x\geqq 0$ , the opposite inequality is valid in this case. We also see, in the same way, that inequality (60) also remains when $p<0$ , provided that the zeros (1), and therefore also the numbers (45), (46), are all positive.
19. - Let us make one last application of inequality (59). The function $\log x$ is concave for $x>0$ . It follows that if the zeros (1) are positive we have the inequality
(61) $\sqrt[n]{x_{1}^{\prime}x_{2}^{\prime}\ldots x_{n}^{\prime}}\geqq\sqrt[n-1]{y_{1}^{\prime}y_{2}^{\prime}\ldots y_{n-1}^{\prime}}\quad(n>2)$ .

This inequality also holds when (1) are only non-negative. The hypothesis $x_{n}^{\prime}=0$ required $x_{1}=x_{2}=\ldots=x_{n-1}=0$ SO $y_{1}=y_{2}=\ldots y_{n=2}=0$ and consequently $y_{n-1}^{\prime}=0$ . It follows that in (61) equality only holds if either all the zeros (1) are equal or the $n-1$ first ones are zero.

Either $f(x)$ a polynomial of degree $n>2$ with all real zeros and $x_{s,1},x_{s,2},\ldots,x_{s,n-s}$ the zeros of the $s$ .th derivative $f^{(s)}(x)$ . Let us designate by $x_{s,1}^{\prime},x_{s,2}^{\prime},\ldots,x_{s,n-s}^{\prime}$ the arithmetic means
of the zeros $x_{s,r}$ taken $n-s-1$ has $n-s-1$ . Inequality (61) gives us the sequence of inequalities

\begin{gathered}\sqrt[n]{x_{0,1}^{\prime}x_{0,2}^{\prime}\ldots x_{0,n}^{\prime}}\geqq\sqrt[n-1]{x_{1,1}^{\prime}x_{1,2}^{\prime}\ldots x_{1,n-1}^{\prime}}\geqq\\ \geqq\sqrt{x_{2,1}^{\prime}x_{2,2}^{\prime}\ldots x_{2,n-2}^{\prime}}\geqq\cdots\geqq\sqrt{x_{n-2,1}^{\prime}x_{n-2,2}^{\prime}}\end{gathered}

assuming, of course, that the zeros of $f(x)$ are all non-negative.

It is easy to see that for all these inequalities the conditions of equality are the same as for (61).
20. - Finally, let us change the notations a little again. Let us denote by $a_{1},a_{n},\ldots,a_{n}$ the zeros of $f(x)$ . We can easily see that $x_{n-2,1}^{\prime}x_{n-2,2}$ are the zeros of $f^{(n-2)}(x)$ . But, the product of these zeros is equal to the arithmetic mean of the two-by-two products of the numbers $a_{i}$ . So we get the following property.

THEOREM 14. - If $a_{1},a_{2},\ldots,a_{n}(n>2)$ are non-negative numbers and if $A_{1},A_{2},\ldots,A_{n}$ are the arithmetic means of these numbers taken $n-1$ has $n-1$ , we have the inequality

\sqrt[n]{A_{1}A_{2}\ldots A_{n}}\geqq\sqrt{\frac{\sum a_{i}a_{j}}{\binom{n}{2}}}

(62)

equality taking place if and only if either all numbers $a_{i}$ are equal, or $n-1$ of these numbers are zero.

Our inequality, for $n=3$ and for $n=4$ , can be written

\sqrt[3]{(a+b)(b+c)(c+a)}\geq\frac{2}{\sqrt{3}}\sqrt{ab+bc+ca}

$\sqrt[4]{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}\geq\frac{\sqrt{3}}{\sqrt{2}}\sqrt{ab+ac+ad+bc+bd+cd}$
$a,b,c,d$ being non-negative numbers.
The interest of inequality (62) consists in the fact that the first member is a superposition of arithmetic means and a geometric mean. If $A,G$ are the arithmetic and geometric means of the numbers $a_{1},a_{2},\ldots,a_{n}$ , we have

A\geqq\sqrt[n]{A_{1}A_{2}\ldots A_{n}}\geqq G

On the other hand we also have

A\geq\sqrt{\frac{\sum a_{i}a_{j}}{\binom{n}{2}}}\geq G

and we see that (62) is an elegant precision of the first of these inequalities.

Iaşi, October 6, 1943.

On certain inequalities between the zeros, assumed to be all real, of a polynomial and those of its derivative

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ON CERTAIN INEQUALITIES BETWEEN THE ZEROES, ALL ASSUMED TO BE REAL, OF A POLYNOMIAL AND THOSE OF ITS DERIVATIVE

$\S 2$ .

But

§ 3.

$\S 4$ .

Lemma 5.-For the inequality

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On certain inequalities between the zeros, assumed to be all real, of a polynomial and those of its derivative

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ON CERTAIN INEQUALITIES BETWEEN THE ZEROES, ALL ASSUMED TO BE REAL, OF A POLYNOMIAL AND THOSE OF ITS DERIVATIVE

§​2\S 2.

But

§ 3.

§​4\S 4.

Lemma 5.-For the inequality

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$\S 2$ .

$\S 4$ .