On an inequality

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T. Popoviciu, Sur une inegalité, Mathematica, 23 (1947-1948), pp. 127-128 (in French).

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Mathematica

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[MR0027024]

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ON AN INEQUALITY

BY

TIBERIU POPOVICIU

Requested on April 27, 1948
In the "New Annals of Mathematics" ( 1 ) ( 1 ) ^((1)){ }^{(1)}(1)LA Le Cointe demonstrates that if has 1 ′ , has 2 ′ , … , has 2 m ′ has 1 ′ , has 2 ′ , … , has 2 m ′ a_(1)^('),a_(2)^('),dots,a_(2m)^(')a_{1}^{\prime}, a_{2}^{\prime}, \ldots, a_{2 m}^{\prime}has1′,has2′,…,has2m′are the 2 m 2 m 2m2 m2mpositive numbers has 1 , has 2 , … , has 2 m has 1 , has 2 , … , has 2 m a_(1),a_(2),dots,a_(2m)a_{1}, a_{2}, \ldots, a_{2 m}has1,has2,…,has2marranged in non-descending (or non-ascending) order, we have
has 1 ′ has 2 ′ … has m ′ + has m + 1 ′ has m + 2 ′ … has 2 m ′ ≧ has 1 has 2 … has m + has m + 1 has m + 2 … has 2 m has 1 ′ has 2 ′ … has m ′ + has m + 1 ′ has m + 2 ′ … has 2 m ′ ≧ has 1 has 2 … has m + has m + 1 has m + 2 … has 2 m a_(1)^(')a_(2)^(')dotsa_(m)^(')+a_(m+1)^(')a_(m+2)^(')dotsa_(2m)^(') >= a_(1)a_(2)dotsa_(m)+a_(m+1)a_(m+2)dotsa_(2m)a_{1}^{\prime} a_{2}^{\prime} \ldots a_{m}^{\prime}+a_{m+1}^{\prime} a_{m+2}^{\prime} \ldots a_{2 m}^{\prime} \geqq a_{1} a_{2} \ldots a_{m}+a_{m+1} a_{m+2} \ldots a_{2m}has1′has2′…hasm′+hasm+1′hasm+2′…has2m′≧has1has2…hasm+hasm+1hasm+2…has2m
We will give a generalization of this property by demonstrating the
Theorem. - Let φ ( x ) , ψ ( x ) φ ( x ) , ψ ( x ) varphi(x),psi(x)\varphi(x), \psi(x)φ(x),ψ(x)two continuous and increasing functions in the interval. ( has , b ) ( has , b ) (a,b)(a, b)(has,b)And φ − 1 , ψ − 1 φ − 1 , ψ − 1 varphi^(-1),psi^(-1)\varphi^{-1}, \psi^{-1}φ−1,ψ−1their inverse functions. Let us set
M ( x 1 , x 2 , … , x n ) = φ − 1 ( ∑ i = 1 n φ ( x i ) n ) N ( x 1 , x 2 , … , x m ) = ψ − 1 ( ∑ i = 1 m ψ ( x i ) m ) M x 1 , x 2 , … , x n = φ − 1 ∑ i = 1 n   φ x i n N x 1 , x 2 , … , x m = ψ − 1 ∑ i = 1 m   ψ x i m {:[M(x_(1),x_(2),dots,x_(n))=varphi^(-1)((sum_(i=1)^(n)varphi(x_(i)))/(n))],[N(x_(1),x_(2),dots,x_(m))=psi^(-1)((sum_(i=1)^(m)psi(x_(i)))/(m))]:}\begin{aligned} & \mathrm{M}\left(x_{1}, x_{2}, \ldots, x_{n}\right)=\varphi^{-1}\left(\frac{\sum_{i=1}^{n} \varphi\left(x_{i}\right)}{n}\right) \\ & \mathrm{N}\left(x_{1}, x_{2}, \ldots, x_{m}\right)=\psi^{-1}\left(\frac{\sum_{i=1}^{m} \psi\left(x_{i}\right)}{m}\right) \end{aligned}M(x1,x2,…,xn)=φ−1(∑i=1nφ(xi)n)N(x1,x2,…,xm)=ψ−1(∑i=1mψ(xi)m)
Let's still be has 1 , has 2 , … , has n m n m has 1 , has 2 , … , has n m n m a_(1),a_(2),dots,a_(nm)nma_{1}, a_{2}, \ldots, a_{nm} nmhas1,has2,…,hasnmnmnumbers belonging to the interval ( has , b ) ( has , b ) (a,b)(a, b)(has,b)And has 1 ′ , has 2 ′ , … , has n m ′ has 1 ′ , has 2 ′ , … , has n m ′ a_(1)^('),a_(2)^('),dots,a_(nm)^(')a_{1}^{\prime}, a_{2}^{\prime}, \ldots, a_{nm}^{\prime}has1′,has2′,…,hasnm′these same numbers arranged in non-increasing (or non-descending) order.
Then we have inequality
(1)
M ( N ( has 1 ′ , has 2 ′ , … , has m ′ ) , N ( has m + 1 ′ , has m + 2 ′ , … , has 2 m ′ ) , … N ( has ( n − 1 ) m + 1 ′ , has ( n − 1 ) m + 2 ′ , … , has n m ′ ) ) ≧ resp . ≦ M ( N ( has 1 , has 2 , … , has m ) , N ( has m + 1 , has m + 2 , … , has 2 m ) , … , N ( has ( n − 1 ) m + 1 , has ( n − 1 ) m + 2 , … , has n m ) ) M N has 1 ′ , has 2 ′ , … , has m ′ , N has m + 1 ′ , has m + 2 ′ , … , has 2 m ′ , … N has ( n − 1 ) m + 1 ′ , has ( n − 1 ) m + 2 ′ , … , has n m ′ ≧ resp . ≦ M N has 1 , has 2 , … , has m , N has m + 1 , has m + 2 , … , has 2 m , … , N has ( n − 1 ) m + 1 , has ( n − 1 ) m + 2 , … , has n m {:[M((N)(a_(1)^('),a_(2)^('),dots,a_(m)^(')),N(a_(m+1)^('),a_(m+2)^('),dots,a_(2m)^(')),dots:}],[{:N(a_((n-1)m+1)^('),a_((n-1)m+2)^('),dots,a_(nm)^('))) >= resp. <= ],[M((N)(a_(1),a_(2),dots,a_(m)),N(a_(m+1),a_(m+2),dots,a_(2m)),dots,:}],[{:N(a_((n-1)m+1),a_((n-1)m+2),dots,a_(nm)))]:}\begin{gathered} \mathrm{M}\left(\mathrm{~N}\left(a_{1}^{\prime}, a_{2}^{\prime}, \ldots, a_{m}^{\prime}\right), \mathrm{N}\left(a_{m+1}^{\prime}, a_{m+2}^{\prime}, \ldots, a_{2 m}^{\prime}\right), \ldots\right. \\ \left.\mathrm{N}\left(a_{(n-1) m+1}^{\prime}, a_{(n-1) m+2}^{\prime}, \ldots, a_{n m}^{\prime}\right)\right) \geqq \mathrm{resp} . \leqq \\ \mathrm{M}\left(\mathrm{~N}\left(a_{1}, a_{2}, \ldots, a_{m}\right), \mathrm{N}\left(a_{m+1}, a_{m+2}, \ldots, a_{2 m}\right), \ldots,\right. \\ \left.\mathrm{N}\left(a_{(n-1) m+1}, a_{(n-1) m+2}, \ldots, a_{n m}\right)\right) \end{gathered}M( N(has1′,has2′,…,hasm′),N(hasm+1′,hasm+2′,…,has2m′),…N(has(n−1)m+1′,has(n−1)m+2′,…,hasnm′))≧resp.≦M( N(has1,has2,…,hasm),N(hasm+1,hasm+2,…,has2m),…,N(has(n−1)m+1,has(n−1)m+2,…,hasnm))
depending on the function φ ( ψ − 1 ) φ ψ − 1 varphi(psi^(-1))\varphi\left(\psi^{-1}\right)φ(ψ−1)is non-concave resp. non-convex.
( 1 1 ^(1){ }^{1}1) T. 2. 372-374 (1843).
Let us designate by f f fffthe function φ ( ψ − 1 ) φ ψ − 1 varphi(psi^(-1))\varphi\left(\psi^{-1}\right)φ(ψ−1)and let us suppose, to fix the ideas, that this function is non-concave. The functions φ − 1 , ψ − 1 φ − 1 , ψ − 1 varphi^(-1),psi^(-1)\varphi^{-1}, \psi^{-1}φ−1,ψ−1being also increasing, the inequality to be demonstrated amounts to
∑ i = 1 n f ( ∑ j = 1 m ψ ( a ( i − 1 ) m + j ′ ) m m ) ≧ ∑ i = 1 n f ( ∑ j = 1 m ψ ( a ( i − 1 ) m + j ) m m ) ∑ i = 1 n &hairsp; f ∑ j = 1 m &hairsp; ψ a ( i − 1 ) m + j ′ m m ≧ ∑ i = 1 n &hairsp; f ∑ j = 1 m &hairsp; ψ a ( i − 1 ) m + j m m sum_(i=1)^(n)f((sum_(j=1)^(m)(psi(a_((i-1)m+j)^(')))/(m))/(m)) >= sum_(i=1)^(n)f((sum_(j=1)^(m)(psi(a_((i-1)m+j)))/(m))/(m))\sum_{i=1}^{n} f\left(\frac{\sum_{j=1}^{m} \frac{\psi\left(a_{(i-1) m+j}^{\prime}\right)}{m}}{m}\right) \geqq \sum_{i=1}^{n} f\left(\frac{\sum_{j=1}^{m} \frac{\psi\left(a_{(i-1) m+j}\right)}{m}}{m}\right)∑i=1nf(∑I=1mψ(has(i−1)m+I′)mm)≧∑i=1nf(∑I=1mψ(has(i−1)m+I)mm)
which is a consequence of the following property of Messrs. GH Hardy, JE Littlewood and G. Polya ( 2 ) 2 (^(2))\left({ }^{2}\right)(2).
If f ( x ) f ( x ) f(x)f(x)f(x)is a non-concave function and if ( x i ≥ x i + 1 , y i ≧ y i + 1 , i = 1 , 2 , … , n − 1 ) x i ≥ x i + 1 , y i ≧ y i + 1 , i = 1 , 2 , … , n − 1 ) (x_(i) >= x_(i+1),y_(i) >= y_(i+1),i=:}1,2,dots,n-1)\left(x_{i} \geq x_{i+1}, y_{i} \geqq y_{i+1}, i=\right. 1,2, \ldots, n-1)(xi≥xi+1,yi≧yi+1,i=1,2,…,n−1)
x 1 + x 2 + ⋯ + x i ≧ y 1 + y 2 + ⋯ + y i , i = 1 , 2 , … , n − 1 , x 1 + x 2 + ⋯ + x n = y 1 + y 2 + ⋯ + y n x 1 + x 2 + ⋯ + x i ≧ y 1 + y 2 + ⋯ + y i , i = 1 , 2 , … , n − 1 , x 1 + x 2 + ⋯ + x n = y 1 + y 2 + ⋯ + y n {:[x_(1)+x_(2)+cdots+x_(i) >= y_(1)+y_(2)+cdots+y_(i)","quad i=1","2","dots","n-1","],[x_(1)+x_(2)+cdots+x_(n)=y_(1)+y_(2)+cdots+y_(n)]:}\begin{aligned} & x_{1}+x_{2}+\cdots+x_{i} \geqq y_{1}+y_{2}+\cdots+y_{i}, \quad i=1,2, \ldots, n-1, \\ & x_{1}+x_{2}+\cdots+x_{n}=y_{1}+y_{2}+\cdots+y_{n} \end{aligned}x1+x2+⋯+xi≧y1+y2+⋯+yi,i=1,2,…,n−1,x1+x2+⋯+xn=y1+y2+⋯+yn
we have inequality
∑ i = 1 n f ( x i ) ≧ ∑ i = 1 n f ( y i ) ∑ i = 1 n &hairsp; f x i ≧ ∑ i = 1 n &hairsp; f y i sum_(i=1)^(n)f(x_(i)) >= sum_(i=1)^(n)f(y_(i))\sum_{i=1}^{n} f\left(x_{i}\right) \geqq \sum_{i=1}^{n} f\left(y_{i}\right)∑i=1nf(xi)≧∑i=1nf(yi)
In the case where the function φ ( ψ − 1 ) φ ψ − 1 varphi(psi^(-1))\varphi\left(\psi^{-1}\right)φ(ψ−1)is convex (resp. concave) it is easy to find the cases where the equality is valid in (1).
The property of Le Cointe corresponds to the case where n = 2 , φ = x , ψ = log ⁡ x n = 2 , φ = x , ψ = log ⁡ x n=2,varphi=x,psi=log xn=2, \varphi=x, \psi=\log xn=2,φ=x,ψ=log⁡x.
By particularizing the functions φ , ψ φ , ψ varphi,psi\varphi, \psiφ,ψwe find various specific statements.

  1. ( 2 ) ( 2 ) ^((2)){ }^{(2)}(2)„Some simple inequalities satisfied by convex function", Messenger of Math., 58, 145-152 (1929). See also "Inequalities", p. 89.
1948

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