Abstract
We consider weighted arithmetic means as, for example \(\alpha G+\left(1-\alpha \right) C\), with \(\alpha \in \left( 0,1\right) ,GC\) being the geometric and anti-harmonic means, and we find the range of values of \(\alpha\) for which the weighted mean is still greater or less than some suitable means, in this case the arithmetic and H\”{o}lder ones
Authors
Mira-Cristiana Anisiu
Tiberiu Popoviciu, Institute of Numerical Analysis, Romanian Academy, Romania
Valeriu Anisiu
Babeş-Bolyai University, Cluj-Napoca, Romania
Keywords
inequalities; asymptotic developments; symbolic algebra
Paper coordinates
M.-C. Anisiu, V. Anisiu, Refinement of some inequalities for means, Rev. Anal. Numér. Théor. Approx. 35 (1) (2006), 5-10, https://doi.org/10.33993/jnaat351-1005
About this paper
Journal
Journal of Numerical Analysis and Approximation Theory
Publisher Name
Romanian Academy
Print ISSN
2457-6794
Online ISSN
E 2501-059X
google scholar link
[1] Bullen, P. S., Handbook of Means and Their Inequalities, Series: Mathematics and Its Applications, vol. 560, 2nd ed., Kluwer Academic Publishers Group, Dordrecht, 2003.
[2] Ivan, M. and Raşa, I., Some inequalities for means, Tiberiu Popoviciu Itinerant Seminar of Functional Equations, Approximation and Convexity, Cluj-Napoca, May 23-29, pp. 99-102, 2000.
REVUE D'ANALYSE NUMÉRIQUE ET DE THÉORIE DE L'APPROXIMATION
Rev. Anal. Numér. Théor. Approx., vol. 35 (2006) no. 1, pp. 5-10
ictp.acad.ro/jnaat
ictp.acad.ro/jnaat
REFINEMENT OF SOME INEQUALITIES FOR MEANS
Abstract
We consider weighted arithmetic means as, for example, alpha G+(1-alpha)C\alpha G+ (1-\alpha) C, with alpha in(0,1),G,C\alpha \in(0,1), G, C being the geometric and anti-harmonic means, and we find the range of values of alpha\alpha for which the weighted mean is still greater or less than some suitable means, in this case the arithmetic and Hölder ones.
MSC 2000. 26D15, 26D99.
Keywords. Inequalities, asymptotic developments, symbolic algebra.
Keywords. Inequalities, asymptotic developments, symbolic algebra.
1. INTRODUCTION
It is well-known that the classical means, namely the arithmetic, geometric and harmonic ones,
A=(a+b)/(2),quad G=sqrt(ab),quad H=(2ab)/(a+b)A=\frac{a+b}{2}, \quad G=\sqrt{a b}, \quad H=\frac{2 a b}{a+b}
satisfy the inequalities
H < G < A,H<G<A,
for 0 < a < b0<a<b. As in [2], we shall consider some other means, like:
- the Hölder and the anti-harmonic mean
Q=((a^(2)+b^(2))/(2))^(1//2),quad C=(a^(2)+b^(2))/(a+b)Q=\left(\frac{a^{2}+b^{2}}{2}\right)^{1 / 2}, \quad C=\frac{a^{2}+b^{2}}{a+b}
- the Pólya & Szegő logarithmic mean, the exponential (or identric), and the weighted geometric mean
L=(b-a)/(ln b-ln a),quad I=(1)/(e)((b^(b))/(a^(a)))^(1//(b-a)),quad S=(a^(a)b^(b))^(1//(a+b))L=\frac{b-a}{\ln b-\ln a}, \quad I=\frac{1}{\mathrm{e}}\left(\frac{b^{b}}{a^{a}}\right)^{1 /(b-a)}, \quad S=\left(a^{a} b^{b}\right)^{1 /(a+b)}
An exhaustive bibliography and a full treatment of the topic can be found in [1.
In [2], beside the known inequalities
{:(1)H < G < L < I < A < Q < S < C:}\begin{equation*}
H<G<L<I<A<Q<S<C \tag{1}
\end{equation*}
the authors present the following relation between some means
{:(2)(G+Q)/(2) < A < (G+C)/(2) < Q < (A+C)/(2) < S.:}\begin{equation*} \frac{G+Q}{2}<A<\frac{G+C}{2}<Q<\frac{A+C}{2}<S . \tag{2} \end{equation*}
{:(2)(G+Q)/(2) < A < (G+C)/(2) < Q < (A+C)/(2) < S.:}\begin{equation*} \frac{G+Q}{2}<A<\frac{G+C}{2}<Q<\frac{A+C}{2}<S . \tag{2} \end{equation*}
We can add to (2) the following inequalities
{:(3)L < (G+A)/(2) < (G+Q)/(2).:}\begin{equation*}
L<\frac{G+A}{2}<\frac{G+Q}{2} . \tag{3}
\end{equation*}
The first one will follow from Proposition 1 of the next section (the second is obvious in view of (1)).
In section 2 we shall consider weighted arithmetic means as, for example, alpha G+(1-alpha)C\alpha G+(1-\alpha) C, with alpha in(0,1)\alpha \in(0,1) instead of (G+C)//2(G+C) / 2, and we shall find the range of values of alpha\alpha for which the weighted mean is still greater or less than its neighbours in (2) or (3).
In section 3 we prove that alpha A+(1-alpha)S > Q\alpha A+(1-\alpha) S>Q if and only if alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2}. As a consequence, there are numbers 0 < a < b0<a<b for which (A+S)//2 > Q(A+S) / 2>Q, while for other pairs of numbers (A+S)//2 < Q(A+S) / 2<Q.
2. REFINED INEQUALITIES
Let us denote t=b//a,t > 1t=b / a, t>1. It it obvious that, if M(a,b)M(a, b) is any mean, it suffices to prove the inequalities in (1), (2) or (3) for M(1,t)M(1, t). We shall write from now on M(t)M(t) instead of M(1,t)M(1, t).
Proposition 1. 1. L(t) < alpha G(t)+(1-alpha)A(t),AA t > 1L(t)<\alpha G(t)+(1-\alpha) A(t), \forall t>1 if and only if alpha <= (2)/(3)\alpha \leq \frac{2}{3};
2. alpha G(t)+(1-alpha)Q(t) < A(t),AA t > 1\alpha G(t)+(1-\alpha) Q(t)<A(t), \forall t>1 if and only if alpha >= (1)/(2)\alpha \geq \frac{1}{2}.
2. alpha G(t)+(1-alpha)Q(t) < A(t),AA t > 1\alpha G(t)+(1-\alpha) Q(t)<A(t), \forall t>1 if and only if alpha >= (1)/(2)\alpha \geq \frac{1}{2}.
Proof. 1. We have L(t) < alpha G(t)+(1-alpha)A(t)L(t)<\alpha G(t)+(1-\alpha) A(t) if and only if
(A(t)-L(t))/(A(t)-G(t)) > alpha\frac{A(t)-L(t)}{A(t)-G(t)}>\alpha
We denote
{:(4)f_(11)(t)=(A(t)-L(t))/(A(t)-G(t))=((t+1)ln t-2(t-1))/((t+1-2sqrtt)ln t):}\begin{equation*}
f_{11}(t)=\frac{A(t)-L(t)}{A(t)-G(t)}=\frac{(t+1) \ln t-2(t-1)}{(t+1-2 \sqrt{t}) \ln t} \tag{4}
\end{equation*}
The limits at 1 and oo\infty are lim_(t rarr1)f_(11)(t)=2//3\lim _{t \rightarrow 1} f_{11}(t)=2 / 3 and lim_(t rarr oo)f_(11)(t)=1\lim _{t \rightarrow \infty} f_{11}(t)=1. We evaluate f_(11)(t)-2//3f_{11}(t)-2 / 3 and show that it is positive. The denominator is obviously positive; we substitute u=sqrttu=\sqrt{t} in the numerator and obtain
f(u)=(u^(2)+4u+1)ln u-3u^(2)+3.f(u)=\left(u^{2}+4 u+1\right) \ln u-3 u^{2}+3 .
We have f(1)=f^(')(1)=f^('')(1)=0f(1)=f^{\prime}(1)=f^{\prime \prime}(1)=0 and f^(''')(u)=2(u-1)^(2)//u^(3) > 0f^{\prime \prime \prime}(u)=2(u-1)^{2} / u^{3}>0 for u > 1u>1, hence f_(11)(t) > 2//3f_{11}(t)>2 / 3 for t > 1t>1.
2. Let us consider for t > 1t>1, the function
2. Let us consider for t > 1t>1, the function
{:(5)f_(12)(t)=(Q(t)-A(t))/(Q(t)-G(t))=(sqrt(2(t^(2)+1))-(t+1))/(sqrt(2(t^(2)+1))-2sqrtt)=1-(sqrt(t^(2)+1)+sqrt(2t))/(sqrt2(sqrtt+1)^(2)).:}\begin{equation*}
f_{12}(t)=\frac{Q(t)-A(t)}{Q(t)-G(t)}=\frac{\sqrt{2\left(t^{2}+1\right)}-(t+1)}{\sqrt{2\left(t^{2}+1\right)}-2 \sqrt{t}}=1-\frac{\sqrt{t^{2}+1}+\sqrt{2 t}}{\sqrt{2}(\sqrt{t}+1)^{2}} . \tag{5}
\end{equation*}
We have f_(12)(t) < 1//2f_{12}(t)<1 / 2, since sqrt(t^(2)+1)+sqrt(2t) > sqrt2//2(sqrtt+1)^(2)<=>sqrt(t^(2)+1) > sqrt2//2(t+1)<=>(t-1)^(2) > 0\sqrt{t^{2}+1}+\sqrt{2 t}>\sqrt{2} / 2(\sqrt{t}+1)^{2} \Leftrightarrow \sqrt{t^{2}+1}> \sqrt{2} / 2(t+1) \Leftrightarrow(t-1)^{2}>0, and the conclusion follows since lim_(t rarr1)f_(12)(t)=1//2\lim _{t \rightarrow 1} f_{12}(t)= 1 / 2.
Proposition 2. 1. A(t) < alpha G(t)+(1-alpha)C(t),AA t > 1A(t)<\alpha G(t)+(1-\alpha) C(t), \forall t>1 if and only if alpha <= (1)/(2);\alpha \leq \frac{1}{2} ;
2. alpha G(t)+(1-alpha)C(t) < Q(t),AA t > 1\alpha G(t)+(1-\alpha) C(t)<Q(t), \forall t>1 if and only if alpha >= alpha_(0)\alpha \geq \alpha_{0}, where alpha_(0)=f_(22)(sqrt(u_(0)))=0.3471574308 dots\alpha_{0}=f_{22}\left(\sqrt{u_{0}}\right)=0.3471574308 \ldots, with u_(0)u_{0} the unique root of (8) which is greater than 1 , and f_(22)f_{22} defined in (7).
2. alpha G(t)+(1-alpha)C(t) < Q(t),AA t > 1\alpha G(t)+(1-\alpha) C(t)<Q(t), \forall t>1 if and only if alpha >= alpha_(0)\alpha \geq \alpha_{0}, where alpha_(0)=f_(22)(sqrt(u_(0)))=0.3471574308 dots\alpha_{0}=f_{22}\left(\sqrt{u_{0}}\right)=0.3471574308 \ldots, with u_(0)u_{0} the unique root of (8) which is greater than 1 , and f_(22)f_{22} defined in (7).
Proof. 1. For t > 1t>1 we define
{:(6)f_(21)(t)=(C(t)-A(t))/(C(t)-G(t))=((t-1)^(2))/(2(t^(2)+1-sqrtt(t+1))):}\begin{equation*}
f_{21}(t)=\frac{C(t)-A(t)}{C(t)-G(t)}=\frac{(t-1)^{2}}{2\left(t^{2}+1-\sqrt{t}(t+1)\right)} \tag{6}
\end{equation*}
It follows that f_(21)(t) > 1//2f_{21}(t)>1 / 2, since
f_(21)(t)-(1)/(2)=(sqrtt(sqrtt-1)^(2))/(2(t^(2)+1-sqrtt(t+1)))=(sqrtt)/(2(t+sqrtt+1)) > 0f_{21}(t)-\frac{1}{2}=\frac{\sqrt{t}(\sqrt{t}-1)^{2}}{2\left(t^{2}+1-\sqrt{t}(t+1)\right)}=\frac{\sqrt{t}}{2(t+\sqrt{t}+1)}>0
The infimum of f_(21)f_{21} on ( 1,oo1, \infty ) is precisely 1//21 / 2, because lim_(t rarr1)f_(21)(t)=1//2\lim _{t \rightarrow 1} f_{21}(t)=1 / 2.
2. We consider now
2. We consider now
{:(7)f_(22)(t)=(C(t)-Q(t))/(C(t)-G(t))=(1)/(2)(2(t^(2)+1)-(t+1)sqrt(2(t^(2)+1)))/(t^(2)+1-(t+1)sqrtt):}\begin{equation*}
f_{22}(t)=\frac{C(t)-Q(t)}{C(t)-G(t)}=\frac{1}{2} \frac{2\left(t^{2}+1\right)-(t+1) \sqrt{2\left(t^{2}+1\right)}}{t^{2}+1-(t+1) \sqrt{t}} \tag{7}
\end{equation*}
and obtain lim_(t rarr1)f_(22)(t)=1//3\lim _{t \rightarrow 1} f_{22}(t)=1 / 3 and lim_(t rarr oo)f_(22)(t)=(2-sqrt2)//2\lim _{t \rightarrow \infty} f_{22}(t)=(2-\sqrt{2}) / 2. In order to find the maximum of f_(22)f_{22} we calculate the roots of the derivative of f_(22)f_{22}. Denoting by u=sqrttu=\sqrt{t}, we obtain a unique root in ( 1,oo1, \infty ) of f_(22)^(')f_{22}^{\prime} from
{:(8)u^(8)-8u^(5)-10u^(4)-8u^(3)+1=0:}\begin{equation*}
u^{8}-8 u^{5}-10 u^{4}-8 u^{3}+1=0 \tag{8}
\end{equation*}
which is u_(0)=2.3859965175 dotsu_{0}=2.3859965175 \ldots, for which f_(22)(sqrt(u_(0)))=0.3471574308 dotsf_{22}\left(\sqrt{u_{0}}\right)=0.3471574308 \ldots. The function f_(22)f_{22} will increase up to f_(22)(sqrt(u_(0)))=alpha_(0)f_{22}\left(\sqrt{u_{0}}\right)=\alpha_{0} and then will decrease to (2-sqrt2)//2(2-\sqrt{2}) / 2.
Lemma 3. For t > 1t>1, the following inequality holds
{:(9)t^((t)/(t+1)) > t-ln t:}\begin{equation*}
t^{\frac{t}{t+1}}>t-\ln t \tag{9}
\end{equation*}
Proof. The inequality (9) is equivalent to
(t)/(t+1)ln t > ln(t-ln t)\frac{t}{t+1} \ln t>\ln (t-\ln t)
We consider the function
k(t)=ln(t-ln t)-(t-1)/(t)ln t,quad t > 1k(t)=\ln (t-\ln t)-\frac{t-1}{t} \ln t, \quad t>1
with
k^(')(t)=((ln t-1)ln t)/(t^(2)(t-ln t))k^{\prime}(t)=\frac{(\ln t-1) \ln t}{t^{2}(t-\ln t)}
It has lim_(t rarr1)k(t)=0,lim_(t rarr oo)k(t)=0\lim _{t \rightarrow 1} k(t)=0, \lim _{t \rightarrow \infty} k(t)=0 and a minimum at t_(0)=et_{0}=\mathrm{e}. It follows that k(t) < 0k(t)<0 on (1,oo)(1, \infty), hence ((t-1)//t)ln t > ln(t-ln t)((t-1) / t) \ln t>\ln (t-\ln t). It follows that
(t)/(t+1)ln t > (t-1)/(t)ln t > ln(t-ln t)\frac{t}{t+1} \ln t>\frac{t-1}{t} \ln t>\ln (t-\ln t)
Now we can prove
Proposition 4. 1. Q(t) < alpha A(t)+(1-alpha)C(t),AA t > 1Q(t)<\alpha A(t)+(1-\alpha) C(t), \forall t>1 if and only if alpha <= (1)/(2)\alpha \leq \frac{1}{2};
2. alpha A(t)+(1-alpha)C(t) < S(t),AA t > 1\alpha A(t)+(1-\alpha) C(t)<S(t), \forall t>1 if and only if alpha >= (1)/(2)\alpha \geq \frac{1}{2}.
Proposition 4. 1. Q(t) < alpha A(t)+(1-alpha)C(t),AA t > 1Q(t)<\alpha A(t)+(1-\alpha) C(t), \forall t>1 if and only if alpha <= (1)/(2)\alpha \leq \frac{1}{2};
2. alpha A(t)+(1-alpha)C(t) < S(t),AA t > 1\alpha A(t)+(1-\alpha) C(t)<S(t), \forall t>1 if and only if alpha >= (1)/(2)\alpha \geq \frac{1}{2}.
Proof. 1. Let us consider, for t > 1t>1
{:(10)f_(41)(t)=(C(t)-Q(t))/(C(t)-A(t))=(2(t^(2)+1)-(t+1)sqrt(2(t^(2)+1)))/((t-1)^(2)):}\begin{equation*}
f_{41}(t)=\frac{C(t)-Q(t)}{C(t)-A(t)}=\frac{2\left(t^{2}+1\right)-(t+1) \sqrt{2\left(t^{2}+1\right)}}{(t-1)^{2}} \tag{10}
\end{equation*}
It follows that
f_(41)(t)-(1)/(2)=(3(t^(2)+1)+2t-2(t+1)sqrt(2(t^(2)+1)))/(2(t-1)^(2)) >= 0f_{41}(t)-\frac{1}{2}=\frac{3\left(t^{2}+1\right)+2 t-2(t+1) \sqrt{2\left(t^{2}+1\right)}}{2(t-1)^{2}} \geq 0
because (3(t^(2)+1)+2t)^(2)-8(t+1)^(2)(t^(2)+1) >= 0<=>(t-1)^(4)\left(3\left(t^{2}+1\right)+2 t\right)^{2}-8(t+1)^{2}\left(t^{2}+1\right) \geq 0 \Leftrightarrow(t-1)^{4}. We have lim_(t rarr1)f_(41)(t)=1//2\lim _{t \rightarrow 1} f_{41}(t)=1 / 2, hence this is the infimum of f_(41)f_{41} on (1,oo)(1, \infty).
2. Finally we define
2. Finally we define
{:(11)f_(42)(t)=(C(t)-S(t))/(C(t)-A(t))=2(t^(2)+1-(t+1)t^((t)/(t+1)))/((t-1)^(2)):}\begin{equation*}
f_{42}(t)=\frac{C(t)-S(t)}{C(t)-A(t)}=2 \frac{t^{2}+1-(t+1) t^{\frac{t}{t+1}}}{(t-1)^{2}} \tag{11}
\end{equation*}
We have
f_(42)(t)-(1)/(2)=(3(t^(2)+1)+2t-4(t+1)t^((t)/(t+1)))/(4(t-1)^(2))f_{42}(t)-\frac{1}{2}=\frac{3\left(t^{2}+1\right)+2 t-4(t+1) t^{\frac{t}{t+1}}}{4(t-1)^{2}}
We consider the function
g(t)=(3(t^(2)+1)+2t)/(4(t+1))-t^((t)/(t+1))g(t)=\frac{3\left(t^{2}+1\right)+2 t}{4(t+1)}-t^{\frac{t}{t+1}}
and the numerator of its derivative
{:(12)g_(1)(t)=4t^((t)/(t+1))(t+1+ln t)+1-3t^(2)-6t:}\begin{equation*}
g_{1}(t)=4 t^{\frac{t}{t+1}}(t+1+\ln t)+1-3 t^{2}-6 t \tag{12}
\end{equation*}
Using the fact that S > QS>Q, i.e., t^(t//(t+1)) > sqrt((t^(2)+1)//2)t^{t /(t+1)}>\sqrt{\left(t^{2}+1\right) / 2}, we obtain that g_(1)(t) > sqrt(2(t^(2)+1))g_(2)(t)g_{1}(t)> \sqrt{2\left(t^{2}+1\right)} g_{2}(t), where g_(2)(t)=2(t+1+ln t)-(3t^(2)+6t-1)//sqrt(2(t^(2)+1))g_{2}(t)=2(t+1+\ln t)-\left(3 t^{2}+6 t-1\right) / \sqrt{2\left(t^{2}+1\right)}. The numerator of g_(2)^(')g_{2}^{\prime} is (t+1)(sqrt(2(t^(2)+1)))^(3)-(3t^(4)+7t^(2)+6t)(t+1)\left(\sqrt{2\left(t^{2}+1\right)}\right)^{3}-\left(3 t^{4}+7 t^{2}+6 t\right) and it is positive on (1,10)(1,10). It follows that g_(2)(t) > g_(2)(1)=0g_{2}(t)>g_{2}(1)=0, therefore g_(1)g_{1} is positive for 1 < t < 101<t<10.
Let us consider now that t >= 10t \geq 10. Using (9) in (12) we obtain that g_(1)(t) > g_(3)(t)=t^(2)-2t+2-(2ln t+1)^(2)g_{1}(t)> g_{3}(t)=t^{2}-2 t+2-(2 \ln t+1)^{2}. For g_(4)(t)=sqrt(t^(2)-2t+2)-2ln t-1g_{4}(t)=\sqrt{t^{2}-2 t+2}-2 \ln t-1, the sign of g_(4)^(')g_{4}^{\prime} is given by t^(2)-t-2sqrt(t^(2)-2t+2)t^{2}-t-2 \sqrt{t^{2}-2 t+2}; but (t^(2)-t)^(2)-4(t^(2)-2t+2)=(t-10)^(4)+38(t-10)^(3)+537(t-10)^(2)+3348(t-10)+7772 > 0\left(t^{2}-t\right)^{2}-4\left(t^{2}-2 t+2\right)= (t-10)^{4}+38(t-10)^{3}+537(t-10)^{2}+3348(t-10)+7772>0 for t >= 10t \geq 10. It follows that g_(3)(t) >= g_(3)(10)=3.45 dots > 0g_{3}(t) \geq g_{3}(10)=3.45 \ldots>0, hence g_(1)g_{1} is positive for t >= 10t \geq 10 too.
In conclusion, g(t) > g(1)=0g(t)>g(1)=0, and f_(42)(t) > 1//2f_{42}(t)>1 / 2 for t > 1t>1; in addition, lim_(t rarr1)f_(42)(t)=1//2\lim _{t \rightarrow 1} f_{42}(t)=1 / 2.
3. ANOTHER INEQUALITY
In general it is not an easy task to find the range where the parameter alpha\alpha may vary. To this aim the Symbolic Algebra Programs as Maple can be of great help.
We consider the following problem:
Find the values of alpha in(0,1)\alpha \in(0,1) for which
Find the values of alpha in(0,1)\alpha \in(0,1) for which
{:(13)Q(t)-alpha A(t)-(1-alpha)S(t) > 0","quad AA t > 1.:}\begin{equation*}
Q(t)-\alpha A(t)-(1-\alpha) S(t)>0, \quad \forall t>1 . \tag{13}
\end{equation*}
In order to find the minimal value of alpha\alpha for which (13) holds, we shall develop asymptotically the function F(t,alpha)=Q(t)-alpha A(t)-(1-alpha)S(t)F(t, \alpha)=Q(t)-\alpha A(t)-(1-\alpha) S(t) for t rarr oot \rightarrow \infty. We denote
{:(14)u=F(t","alpha)=(1)/(2)sqrt(2+2t^(2))-(1)/(2)alpha(1+t)-(1-alpha)t^((t)/(1+t)):}\begin{equation*}
u=F(t, \alpha)=\frac{1}{2} \sqrt{2+2 t^{2}}-\frac{1}{2} \alpha(1+t)-(1-\alpha) t^{\frac{t}{1+t}} \tag{14}
\end{equation*}
Using the command asympt ( u,t,3u, t, 3 ) we obtain the series
{:[(15)((sqrt2)/(2)+(alpha)/(2)-1)t-(alpha)/(2)+(1-alpha)ln(t)],[+(sqrt2//4-(1-alpha)(ln(t)+1//2(ln(t))^(2)))/(t)+O(t^(-2))]:}\begin{align*}
& \left(\frac{\sqrt{2}}{2}+\frac{\alpha}{2}-1\right) t-\frac{\alpha}{2}+(1-\alpha) \ln (t) \tag{15}\\
& +\frac{\sqrt{2} / 4-(1-\alpha)\left(\ln (t)+1 / 2(\ln (t))^{2}\right)}{t}+\mathcal{O}\left(t^{-2}\right)
\end{align*}
(We mention that the term O(t^(-2))\mathcal{O}\left(t^{-2}\right) is used in the sense of Maple; from the point of view of Landau notation it should be O(t^(-2)ln^(3)(t))\mathcal{O}\left(t^{-2} \ln ^{3}(t)\right).)
For u > 0u>0, the condition sqrt2//2+alpha//2-1 >= 0\sqrt{2} / 2+\alpha / 2-1 \geq 0, hence alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2}, is obviously necessary. Now that we know the expected minimal value of alpha\alpha, we state
Theorem 5. The inequality (13) holds if and only if alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2}.
Proof. We shall prove that the inequality holds for alpha=2-sqrt2\alpha=2-\sqrt{2} (hence aa fortiori for alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2} ).
Proof. We shall prove that the inequality holds for alpha=2-sqrt2\alpha=2-\sqrt{2} (hence aa fortiori for alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2} ).
Let us denote f(t)=(sqrt2+1)F(t,2-sqrt2)f(t)=(\sqrt{2}+1) F(t, 2-\sqrt{2}), where FF is given in (14). We have to prove that f(t) > 0f(t)>0 for t > 1t>1. It follows that
f(t)=((sqrt2+1)sqrt(2(t^(2)+1)))/(2)-(sqrt2(t+1))/(2)-t^((t)/(t+1))f(t)=\frac{(\sqrt{2}+1) \sqrt{2\left(t^{2}+1\right)}}{2}-\frac{\sqrt{2}(t+1)}{2}-t^{\frac{t}{t+1}}
We put in the inequality (1+x)^(q) < 1+qx(1+x)^{q}<1+q x, which holds for x > 0,0 < q < 1x>0,0<q<1, x=t-1x=t-1 and q=t//(t+1)q=t /(t+1). It follows that
t^((t)/(t+1)) < (t^(2)+1)/(t+1)t^{\frac{t}{t+1}}<\frac{t^{2}+1}{t+1}
and
f(t) > ((1+sqrt2))/(2(t+1))((t+1)sqrt(2+2t^(2))-sqrt2(t^(2)+2(sqrt2-1)t+1))f(t)>\frac{(1+\sqrt{2})}{2(t+1)}\left((t+1) \sqrt{2+2 t^{2}}-\sqrt{2}\left(t^{2}+2(\sqrt{2}-1) t+1\right)\right)
Let us denote the positive expressions
f_(1)(t)=(t+1)sqrt(2+2t^(2)),quadf_(2)(t)=sqrt2(t^(2)+2(sqrt2-1)t+1);f_{1}(t)=(t+1) \sqrt{2+2 t^{2}}, \quad f_{2}(t)=\sqrt{2}\left(t^{2}+2(\sqrt{2}-1) t+1\right) ;
it follows easily that f_(1)^(2)(t)-f_(2)^(2)(t)=4t(t-1)^(2)f_{1}^{2}(t)-f_{2}^{2}(t)=4 t(t-1)^{2}, therefore f(t) > 0f(t)>0.
REFERENCES
[1] Bullen, P. S., Handbook of Means and Their Inequalities, Series: Mathematics and Its Applications, vol. 560, 2nd ed., Kluwer Academic Publishers Group, Dordrecht, 2003.
[2] Ivan, M. and Raşa, I., Some inequalities for means, Tiberiu Popoviciu Itinerant Seminar of Functional Equations, Approximation and Convexity, Cluj-Napoca, May 23-29, pp. 99-102, 2000.
[2] Ivan, M. and Raşa, I., Some inequalities for means, Tiberiu Popoviciu Itinerant Seminar of Functional Equations, Approximation and Convexity, Cluj-Napoca, May 23-29, pp. 99-102, 2000.
Received by the editors: January 12, 2006.
- *"T. Popoviciu" Institute of Numerical Analysis, P.O. Box 68, 400110 Cluj-Napoca, România, e-mail: mira@math.ubbcluj.ro.
†"Babeş-Bolyai" University, Faculty of Mathematics and Computer Science, 1 Kogălniceanu St., 400084 Cluj-Napoca, România, e-mail: anisiu@math.ubbcluj.ro.