Refinement of some inequalities for means

Mira Anisiu
(original), paper

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

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https://ictp.acad.ro/jnaat/journal/article/view/2006-vol35-no1-art2

About this paper

Journal

Journal of Numerical Analysis and Approximation Theory

Publisher Name

Romanian Academy

DOI

https://doi.org/10.33993/jnaat351-1005

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2457-6794

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E 2501-059X

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[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.

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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

REFINEMENT OF SOME INEQUALITIES FOR MEANS

MIRA-CRISTIANA ANISIU* and VALERIU ANISIU † † ^(†){ }^{\dagger}†

Abstract

We consider weighted arithmetic means as, for example, α G + ( 1 − α ) C α G + ( 1 − α ) C alpha G+(1-alpha)C\alpha G+ (1-\alpha) CαG+(1−α)C, with α ∈ ( 0 , 1 ) , G , C α ∈ ( 0 , 1 ) , G , C alpha in(0,1),G,C\alpha \in(0,1), G, Cα∈(0,1),G,C being the geometric and anti-harmonic means, and we find the range of values of α α alpha\alphaα 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.

1. INTRODUCTION

It is well-known that the classical means, namely the arithmetic, geometric and harmonic ones,
A = a + b 2 , G = a b , H = 2 a b a + b A = a + b 2 , G = a b , H = 2 a b a + b 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}A=a+b2,G=ab,H=2aba+b
satisfy the inequalities
H < G < A , H < G < A , H < G < A,H<G<A,H<G<A,
for 0 < a < b 0 < a < b 0 < a < b0<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 , C = a 2 + b 2 a + b Q = a 2 + b 2 2 1 / 2 , C = a 2 + b 2 a + b 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}Q=(a2+b22)1/2,C=a2+b2a+b
  • the Pólya & Szegő logarithmic mean, the exponential (or identric), and the weighted geometric mean
L = b − a ln ⁡ b − ln ⁡ a , I = 1 e ( b b a a ) 1 / ( b − a ) , S = ( a a b b ) 1 / ( a + b ) L = b − a ln ⁡ b − ln ⁡ a , I = 1 e b b a a 1 / ( b − a ) , S = a a b b 1 / ( a + b ) 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)}L=b−aln⁡b−ln⁡a,I=1e(bbaa)1/(b−a),S=(aabb)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 (1) H < G < L < I < A < Q < S < C {:(1)H < G < L < I < A < Q < S < C:}\begin{equation*} H<G<L<I<A<Q<S<C \tag{1} \end{equation*}(1)H<G<L<I<A<Q<S<C
the authors present the following relation between some means
(2) G + Q 2 < A < G + C 2 < Q < A + C 2 < S . (2) G + Q 2 < A < G + C 2 < Q < A + C 2 < S . {:(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+Q2<A<G+C2<Q<A+C2<S.
We can add to (2) the following inequalities
(3) L < G + A 2 < G + Q 2 . (3) L < G + A 2 < G + Q 2 . {:(3)L < (G+A)/(2) < (G+Q)/(2).:}\begin{equation*} L<\frac{G+A}{2}<\frac{G+Q}{2} . \tag{3} \end{equation*}(3)L<G+A2<G+Q2.
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, α G + ( 1 − α ) C α G + ( 1 − α ) C alpha G+(1-alpha)C\alpha G+(1-\alpha) CαG+(1−α)C, with α ∈ ( 0 , 1 ) α ∈ ( 0 , 1 ) alpha in(0,1)\alpha \in(0,1)α∈(0,1) instead of ( G + C ) / 2 ( G + C ) / 2 (G+C)//2(G+C) / 2(G+C)/2, and we shall find the range of values of α α alpha\alphaα for which the weighted mean is still greater or less than its neighbours in (2) or (3).
In section 3 we prove that α A + ( 1 − α ) S > Q α A + ( 1 − α ) S > Q alpha A+(1-alpha)S > Q\alpha A+(1-\alpha) S>QαA+(1−α)S>Q if and only if α ≥ 2 − 2 α ≥ 2 − 2 alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2}α≥2−2. As a consequence, there are numbers 0 < a < b 0 < a < b 0 < a < b0<a<b0<a<b for which ( A + S ) / 2 > Q ( A + S ) / 2 > Q (A+S)//2 > Q(A+S) / 2>Q(A+S)/2>Q, while for other pairs of numbers ( A + S ) / 2 < Q ( A + S ) / 2 < Q (A+S)//2 < Q(A+S) / 2<Q(A+S)/2<Q.

2. REFINED INEQUALITIES

Let us denote t = b / a , t > 1 t = b / a , t > 1 t=b//a,t > 1t=b / a, t>1t=b/a,t>1. It it obvious that, if M ( a , b ) M ( a , b ) M(a,b)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 ) M(1,t)M(1, t)M(1,t). We shall write from now on M ( t ) M ( t ) M(t)M(t)M(t) instead of M ( 1 , t ) M ( 1 , t ) M(1,t)M(1, t)M(1,t).
Proposition 1. 1. L ( t ) < α G ( t ) + ( 1 − α ) A ( t ) , ∀ t > 1 L ( t ) < α G ( t ) + ( 1 − α ) A ( t ) , ∀ t > 1 L(t) < alpha G(t)+(1-alpha)A(t),AA t > 1L(t)<\alpha G(t)+(1-\alpha) A(t), \forall t>1L(t)<αG(t)+(1−α)A(t),∀t>1 if and only if α ≤ 2 3 α ≤ 2 3 alpha <= (2)/(3)\alpha \leq \frac{2}{3}α≤23;
2. α G ( t ) + ( 1 − α ) Q ( t ) < A ( t ) , ∀ t > 1 α G ( t ) + ( 1 − α ) Q ( t ) < A ( t ) , ∀ t > 1 alpha G(t)+(1-alpha)Q(t) < A(t),AA t > 1\alpha G(t)+(1-\alpha) Q(t)<A(t), \forall t>1αG(t)+(1−α)Q(t)<A(t),∀t>1 if and only if α ≥ 1 2 α ≥ 1 2 alpha >= (1)/(2)\alpha \geq \frac{1}{2}α≥12.
Proof. 1. We have L ( t ) < α G ( t ) + ( 1 − α ) A ( t ) L ( t ) < α G ( t ) + ( 1 − α ) A ( t ) L(t) < alpha G(t)+(1-alpha)A(t)L(t)<\alpha G(t)+(1-\alpha) A(t)L(t)<αG(t)+(1−α)A(t) if and only if
A ( t ) − L ( t ) A ( t ) − G ( t ) > α A ( t ) − L ( t ) A ( t ) − G ( t ) > α (A(t)-L(t))/(A(t)-G(t)) > alpha\frac{A(t)-L(t)}{A(t)-G(t)}>\alphaA(t)−L(t)A(t)−G(t)>α
We denote
(4) f 11 ( t ) = A ( t ) − L ( t ) A ( t ) − G ( t ) = ( t + 1 ) ln ⁡ t − 2 ( t − 1 ) ( t + 1 − 2 t ) ln ⁡ t (4) f 11 ( t ) = A ( t ) − L ( t ) A ( t ) − G ( t ) = ( t + 1 ) ln ⁡ t − 2 ( t − 1 ) ( t + 1 − 2 t ) ln ⁡ t {:(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*}(4)f11(t)=A(t)−L(t)A(t)−G(t)=(t+1)ln⁡t−2(t−1)(t+1−2t)ln⁡t
The limits at 1 and ∞ ∞ oo\infty∞ are lim t → 1 f 11 ( t ) = 2 / 3 lim t → 1   f 11 ( t ) = 2 / 3 lim_(t rarr1)f_(11)(t)=2//3\lim _{t \rightarrow 1} f_{11}(t)=2 / 3limt→1f11(t)=2/3 and lim t → ∞ f 11 ( t ) = 1 lim t → ∞   f 11 ( t ) = 1 lim_(t rarr oo)f_(11)(t)=1\lim _{t \rightarrow \infty} f_{11}(t)=1limt→∞f11(t)=1. We evaluate f 11 ( t ) − 2 / 3 f 11 ( t ) − 2 / 3 f_(11)(t)-2//3f_{11}(t)-2 / 3f11(t)−2/3 and show that it is positive. The denominator is obviously positive; we substitute u = t u = t u=sqrttu=\sqrt{t}u=t in the numerator and obtain
f ( u ) = ( u 2 + 4 u + 1 ) ln ⁡ u − 3 u 2 + 3 . f ( u ) = u 2 + 4 u + 1 ln ⁡ u − 3 u 2 + 3 . 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 .f(u)=(u2+4u+1)ln⁡u−3u2+3.
We have f ( 1 ) = f ′ ( 1 ) = f ′ ′ ( 1 ) = 0 f ( 1 ) = f ′ ( 1 ) = f ′ ′ ( 1 ) = 0 f(1)=f^(')(1)=f^('')(1)=0f(1)=f^{\prime}(1)=f^{\prime \prime}(1)=0f(1)=f′(1)=f′′(1)=0 and f ′ ′ ′ ( u ) = 2 ( u − 1 ) 2 / u 3 > 0 f ′ ′ ′ ( u ) = 2 ( u − 1 ) 2 / u 3 > 0 f^(''')(u)=2(u-1)^(2)//u^(3) > 0f^{\prime \prime \prime}(u)=2(u-1)^{2} / u^{3}>0f′′′(u)=2(u−1)2/u3>0 for u > 1 u > 1 u > 1u>1u>1, hence f 11 ( t ) > 2 / 3 f 11 ( t ) > 2 / 3 f_(11)(t) > 2//3f_{11}(t)>2 / 3f11(t)>2/3 for t > 1 t > 1 t > 1t>1t>1.
2. Let us consider for t > 1 t > 1 t > 1t>1t>1, the function
(5) f 12 ( t ) = Q ( t ) − A ( t ) Q ( t ) − G ( t ) = 2 ( t 2 + 1 ) − ( t + 1 ) 2 ( t 2 + 1 ) − 2 t = 1 − t 2 + 1 + 2 t 2 ( t + 1 ) 2 . (5) f 12 ( t ) = Q ( t ) − A ( t ) Q ( t ) − G ( t ) = 2 t 2 + 1 − ( t + 1 ) 2 t 2 + 1 − 2 t = 1 − t 2 + 1 + 2 t 2 ( t + 1 ) 2 . {:(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*}(5)f12(t)=Q(t)−A(t)Q(t)−G(t)=2(t2+1)−(t+1)2(t2+1)−2t=1−t2+1+2t2(t+1)2.
We have f 12 ( t ) < 1 / 2 f 12 ( t ) < 1 / 2 f_(12)(t) < 1//2f_{12}(t)<1 / 2f12(t)<1/2, since t 2 + 1 + 2 t > 2 / 2 ( t + 1 ) 2 ⇔ t 2 + 1 > 2 / 2 ( t + 1 ) ⇔ ( t − 1 ) 2 > 0 t 2 + 1 + 2 t > 2 / 2 ( t + 1 ) 2 ⇔ t 2 + 1 > 2 / 2 ( t + 1 ) ⇔ ( t − 1 ) 2 > 0 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}>0t2+1+2t>2/2(t+1)2⇔t2+1>2/2(t+1)⇔(t−1)2>0, and the conclusion follows since lim t → 1 f 12 ( t ) = 1 / 2 lim t → 1   f 12 ( t ) = 1 / 2 lim_(t rarr1)f_(12)(t)=1//2\lim _{t \rightarrow 1} f_{12}(t)= 1 / 2limt→1f12(t)=1/2.
Proposition 2. 1. A ( t ) < α G ( t ) + ( 1 − α ) C ( t ) , ∀ t > 1 A ( t ) < α G ( t ) + ( 1 − α ) C ( t ) , ∀ t > 1 A(t) < alpha G(t)+(1-alpha)C(t),AA t > 1A(t)<\alpha G(t)+(1-\alpha) C(t), \forall t>1A(t)<αG(t)+(1−α)C(t),∀t>1 if and only if α ≤ 1 2 ; α ≤ 1 2 ; alpha <= (1)/(2);\alpha \leq \frac{1}{2} ;α≤12;
2. α G ( t ) + ( 1 − α ) C ( t ) < Q ( t ) , ∀ t > 1 α G ( t ) + ( 1 − α ) C ( t ) < Q ( t ) , ∀ t > 1 alpha G(t)+(1-alpha)C(t) < Q(t),AA t > 1\alpha G(t)+(1-\alpha) C(t)<Q(t), \forall t>1αG(t)+(1−α)C(t)<Q(t),∀t>1 if and only if α ≥ α 0 α ≥ α 0 alpha >= alpha_(0)\alpha \geq \alpha_{0}α≥α0, where α 0 = f 22 ( u 0 ) = 0.3471574308 … α 0 = f 22 u 0 = 0.3471574308 … alpha_(0)=f_(22)(sqrt(u_(0)))=0.3471574308 dots\alpha_{0}=f_{22}\left(\sqrt{u_{0}}\right)=0.3471574308 \ldotsα0=f22(u0)=0.3471574308…, with u 0 u 0 u_(0)u_{0}u0 the unique root of (8) which is greater than 1 , and f 22 f 22 f_(22)f_{22}f22 defined in (7).
Proof. 1. For t > 1 t > 1 t > 1t>1t>1 we define
(6) f 21 ( t ) = C ( t ) − A ( t ) C ( t ) − G ( t ) = ( t − 1 ) 2 2 ( t 2 + 1 − t ( t + 1 ) ) (6) f 21 ( t ) = C ( t ) − A ( t ) C ( t ) − G ( t ) = ( t − 1 ) 2 2 t 2 + 1 − t ( t + 1 ) {:(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*}(6)f21(t)=C(t)−A(t)C(t)−G(t)=(t−1)22(t2+1−t(t+1))
It follows that f 21 ( t ) > 1 / 2 f 21 ( t ) > 1 / 2 f_(21)(t) > 1//2f_{21}(t)>1 / 2f21(t)>1/2, since
f 21 ( t ) − 1 2 = t ( t − 1 ) 2 2 ( t 2 + 1 − t ( t + 1 ) ) = t 2 ( t + t + 1 ) > 0 f 21 ( t ) − 1 2 = t ( t − 1 ) 2 2 t 2 + 1 − t ( t + 1 ) = t 2 ( t + t + 1 ) > 0 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)}>0f21(t)−12=t(t−1)22(t2+1−t(t+1))=t2(t+t+1)>0
The infimum of f 21 f 21 f_(21)f_{21}f21 on ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ) is precisely 1 / 2 1 / 2 1//21 / 21/2, because lim t → 1 f 21 ( t ) = 1 / 2 lim t → 1   f 21 ( t ) = 1 / 2 lim_(t rarr1)f_(21)(t)=1//2\lim _{t \rightarrow 1} f_{21}(t)=1 / 2limt→1f21(t)=1/2.
2. We consider now
(7) f 22 ( t ) = C ( t ) − Q ( t ) C ( t ) − G ( t ) = 1 2 2 ( t 2 + 1 ) − ( t + 1 ) 2 ( t 2 + 1 ) t 2 + 1 − ( t + 1 ) t (7) f 22 ( t ) = C ( t ) − Q ( t ) C ( t ) − G ( t ) = 1 2 2 t 2 + 1 − ( t + 1 ) 2 t 2 + 1 t 2 + 1 − ( t + 1 ) t {:(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*}(7)f22(t)=C(t)−Q(t)C(t)−G(t)=122(t2+1)−(t+1)2(t2+1)t2+1−(t+1)t
and obtain lim t → 1 f 22 ( t ) = 1 / 3 lim t → 1   f 22 ( t ) = 1 / 3 lim_(t rarr1)f_(22)(t)=1//3\lim _{t \rightarrow 1} f_{22}(t)=1 / 3limt→1f22(t)=1/3 and lim t → ∞ f 22 ( t ) = ( 2 − 2 ) / 2 lim t → ∞   f 22 ( t ) = ( 2 − 2 ) / 2 lim_(t rarr oo)f_(22)(t)=(2-sqrt2)//2\lim _{t \rightarrow \infty} f_{22}(t)=(2-\sqrt{2}) / 2limt→∞f22(t)=(2−2)/2. In order to find the maximum of f 22 f 22 f_(22)f_{22}f22 we calculate the roots of the derivative of f 22 f 22 f_(22)f_{22}f22. Denoting by u = t u = t u=sqrttu=\sqrt{t}u=t, we obtain a unique root in ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ) of f 22 ′ f 22 ′ f_(22)^(')f_{22}^{\prime}f22′ from
(8) u 8 − 8 u 5 − 10 u 4 − 8 u 3 + 1 = 0 (8) u 8 − 8 u 5 − 10 u 4 − 8 u 3 + 1 = 0 {:(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*}(8)u8−8u5−10u4−8u3+1=0
which is u 0 = 2.3859965175 … u 0 = 2.3859965175 … u_(0)=2.3859965175 dotsu_{0}=2.3859965175 \ldotsu0=2.3859965175…, for which f 22 ( u 0 ) = 0.3471574308 … f 22 u 0 = 0.3471574308 … f_(22)(sqrt(u_(0)))=0.3471574308 dotsf_{22}\left(\sqrt{u_{0}}\right)=0.3471574308 \ldotsf22(u0)=0.3471574308…. The function f 22 f 22 f_(22)f_{22}f22 will increase up to f 22 ( u 0 ) = α 0 f 22 u 0 = α 0 f_(22)(sqrt(u_(0)))=alpha_(0)f_{22}\left(\sqrt{u_{0}}\right)=\alpha_{0}f22(u0)=α0 and then will decrease to ( 2 − 2 ) / 2 ( 2 − 2 ) / 2 (2-sqrt2)//2(2-\sqrt{2}) / 2(2−2)/2.
Lemma 3. For t > 1 t > 1 t > 1t>1t>1, the following inequality holds
(9) t t t + 1 > t − ln ⁡ t (9) t t t + 1 > t − ln ⁡ t {:(9)t^((t)/(t+1)) > t-ln t:}\begin{equation*} t^{\frac{t}{t+1}}>t-\ln t \tag{9} \end{equation*}(9)ttt+1>t−ln⁡t
Proof. The inequality (9) is equivalent to
t t + 1 ln ⁡ t > ln ⁡ ( t − ln ⁡ t ) t t + 1 ln ⁡ t > ln ⁡ ( t − ln ⁡ t ) (t)/(t+1)ln t > ln(t-ln t)\frac{t}{t+1} \ln t>\ln (t-\ln t)tt+1ln⁡t>ln⁡(t−ln⁡t)
We consider the function
k ( t ) = ln ⁡ ( t − ln ⁡ t ) − t − 1 t ln ⁡ t , t > 1 k ( t ) = ln ⁡ ( t − ln ⁡ t ) − t − 1 t ln ⁡ t , t > 1 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>1k(t)=ln⁡(t−ln⁡t)−t−1tln⁡t,t>1
with
k ′ ( t ) = ( ln ⁡ t − 1 ) ln ⁡ t t 2 ( t − ln ⁡ t ) k ′ ( t ) = ( ln ⁡ t − 1 ) ln ⁡ t t 2 ( t − ln ⁡ t ) 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)}k′(t)=(ln⁡t−1)ln⁡tt2(t−ln⁡t)
It has lim t → 1 k ( t ) = 0 , lim t → ∞ k ( t ) = 0 lim t → 1   k ( t ) = 0 , lim t → ∞   k ( t ) = 0 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)=0limt→1k(t)=0,limt→∞k(t)=0 and a minimum at t 0 = e t 0 = e t_(0)=et_{0}=\mathrm{e}t0=e. It follows that k ( t ) < 0 k ( t ) < 0 k(t) < 0k(t)<0k(t)<0 on ( 1 , ∞ ) ( 1 , ∞ ) (1,oo)(1, \infty)(1,∞), hence ( ( t − 1 ) / t ) ln ⁡ t > ln ⁡ ( t − ln ⁡ t ) ( ( t − 1 ) / t ) ln ⁡ t > ln ⁡ ( t − ln ⁡ t ) ((t-1)//t)ln t > ln(t-ln t)((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 ) t t + 1 ln ⁡ t > t − 1 t ln ⁡ t > ln ⁡ ( t − ln ⁡ t ) (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)tt+1ln⁡t>t−1tln⁡t>ln⁡(t−ln⁡t)
Now we can prove
Proposition 4. 1. Q ( t ) < α A ( t ) + ( 1 − α ) C ( t ) , ∀ t > 1 Q ( t ) < α A ( t ) + ( 1 − α ) C ( t ) , ∀ t > 1 Q(t) < alpha A(t)+(1-alpha)C(t),AA t > 1Q(t)<\alpha A(t)+(1-\alpha) C(t), \forall t>1Q(t)<αA(t)+(1−α)C(t),∀t>1 if and only if α ≤ 1 2 α ≤ 1 2 alpha <= (1)/(2)\alpha \leq \frac{1}{2}α≤12;
2. α A ( t ) + ( 1 − α ) C ( t ) < S ( t ) , ∀ t > 1 α A ( t ) + ( 1 − α ) C ( t ) < S ( t ) , ∀ t > 1 alpha A(t)+(1-alpha)C(t) < S(t),AA t > 1\alpha A(t)+(1-\alpha) C(t)<S(t), \forall t>1αA(t)+(1−α)C(t)<S(t),∀t>1 if and only if α ≥ 1 2 α ≥ 1 2 alpha >= (1)/(2)\alpha \geq \frac{1}{2}α≥12.
Proof. 1. Let us consider, for t > 1 t > 1 t > 1t>1t>1
(10) f 41 ( t ) = C ( t ) − Q ( t ) C ( t ) − A ( t ) = 2 ( t 2 + 1 ) − ( t + 1 ) 2 ( t 2 + 1 ) ( t − 1 ) 2 (10) f 41 ( t ) = C ( t ) − Q ( t ) C ( t ) − A ( t ) = 2 t 2 + 1 − ( t + 1 ) 2 t 2 + 1 ( t − 1 ) 2 {:(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*}(10)f41(t)=C(t)−Q(t)C(t)−A(t)=2(t2+1)−(t+1)2(t2+1)(t−1)2
It follows that
f 41 ( t ) − 1 2 = 3 ( t 2 + 1 ) + 2 t − 2 ( t + 1 ) 2 ( t 2 + 1 ) 2 ( t − 1 ) 2 ≥ 0 f 41 ( t ) − 1 2 = 3 t 2 + 1 + 2 t − 2 ( t + 1 ) 2 t 2 + 1 2 ( t − 1 ) 2 ≥ 0 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 0f41(t)−12=3(t2+1)+2t−2(t+1)2(t2+1)2(t−1)2≥0
because ( 3 ( t 2 + 1 ) + 2 t ) 2 − 8 ( t + 1 ) 2 ( t 2 + 1 ) ≥ 0 ⇔ ( t − 1 ) 4 3 t 2 + 1 + 2 t 2 − 8 ( t + 1 ) 2 t 2 + 1 ≥ 0 ⇔ ( t − 1 ) 4 (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}(3(t2+1)+2t)2−8(t+1)2(t2+1)≥0⇔(t−1)4. We have lim t → 1 f 41 ( t ) = 1 / 2 lim t → 1   f 41 ( t ) = 1 / 2 lim_(t rarr1)f_(41)(t)=1//2\lim _{t \rightarrow 1} f_{41}(t)=1 / 2limt→1f41(t)=1/2, hence this is the infimum of f 41 f 41 f_(41)f_{41}f41 on ( 1 , ∞ ) ( 1 , ∞ ) (1,oo)(1, \infty)(1,∞).
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 (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 {:(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*}(11)f42(t)=C(t)−S(t)C(t)−A(t)=2t2+1−(t+1)ttt+1(t−1)2
We have
f 42 ( t ) − 1 2 = 3 ( t 2 + 1 ) + 2 t − 4 ( t + 1 ) t t t + 1 4 ( t − 1 ) 2 f 42 ( t ) − 1 2 = 3 t 2 + 1 + 2 t − 4 ( t + 1 ) t t t + 1 4 ( t − 1 ) 2 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}}f42(t)−12=3(t2+1)+2t−4(t+1)ttt+14(t−1)2
We consider the function
g ( t ) = 3 ( t 2 + 1 ) + 2 t 4 ( t + 1 ) − t t t + 1 g ( t ) = 3 t 2 + 1 + 2 t 4 ( t + 1 ) − t t t + 1 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}}g(t)=3(t2+1)+2t4(t+1)−ttt+1
and the numerator of its derivative
(12) g 1 ( t ) = 4 t t t + 1 ( t + 1 + ln ⁡ t ) + 1 − 3 t 2 − 6 t (12) g 1 ( t ) = 4 t t t + 1 ( t + 1 + ln ⁡ t ) + 1 − 3 t 2 − 6 t {:(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*}(12)g1(t)=4ttt+1(t+1+ln⁡t)+1−3t2−6t
Using the fact that S > Q S > Q S > QS>QS>Q, i.e., t t / ( t + 1 ) > ( t 2 + 1 ) / 2 t t / ( t + 1 ) > t 2 + 1 / 2 t^(t//(t+1)) > sqrt((t^(2)+1)//2)t^{t /(t+1)}>\sqrt{\left(t^{2}+1\right) / 2}tt/(t+1)>(t2+1)/2, we obtain that g 1 ( t ) > 2 ( t 2 + 1 ) g 2 ( t ) g 1 ( t ) > 2 t 2 + 1 g 2 ( t ) g_(1)(t) > sqrt(2(t^(2)+1))g_(2)(t)g_{1}(t)> \sqrt{2\left(t^{2}+1\right)} g_{2}(t)g1(t)>2(t2+1)g2(t), where g 2 ( t ) = 2 ( t + 1 + ln ⁡ t ) − ( 3 t 2 + 6 t − 1 ) / 2 ( t 2 + 1 ) g 2 ( t ) = 2 ( t + 1 + ln ⁡ t ) − 3 t 2 + 6 t − 1 / 2 t 2 + 1 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)}g2(t)=2(t+1+ln⁡t)−(3t2+6t−1)/2(t2+1). The numerator of g 2 ′ g 2 ′ g_(2)^(')g_{2}^{\prime}g2′ is ( t + 1 ) ( 2 ( t 2 + 1 ) ) 3 − ( 3 t 4 + 7 t 2 + 6 t ) ( t + 1 ) 2 t 2 + 1 3 − 3 t 4 + 7 t 2 + 6 t (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)(t+1)(2(t2+1))3−(3t4+7t2+6t) and it is positive on ( 1 , 10 ) ( 1 , 10 ) (1,10)(1,10)(1,10). It follows that g 2 ( t ) > g 2 ( 1 ) = 0 g 2 ( t ) > g 2 ( 1 ) = 0 g_(2)(t) > g_(2)(1)=0g_{2}(t)>g_{2}(1)=0g2(t)>g2(1)=0, therefore g 1 g 1 g_(1)g_{1}g1 is positive for 1 < t < 10 1 < t < 10 1 < t < 101<t<101<t<10.
Let us consider now that t ≥ 10 t ≥ 10 t >= 10t \geq 10t≥10. Using (9) in (12) we obtain that g 1 ( t ) > g 3 ( t ) = t 2 − 2 t + 2 − ( 2 ln ⁡ t + 1 ) 2 g 1 ( t ) > g 3 ( t ) = t 2 − 2 t + 2 − ( 2 ln ⁡ t + 1 ) 2 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}g1(t)>g3(t)=t2−2t+2−(2ln⁡t+1)2. For g 4 ( t ) = t 2 − 2 t + 2 − 2 ln ⁡ t − 1 g 4 ( t ) = t 2 − 2 t + 2 − 2 ln ⁡ t − 1 g_(4)(t)=sqrt(t^(2)-2t+2)-2ln t-1g_{4}(t)=\sqrt{t^{2}-2 t+2}-2 \ln t-1g4(t)=t2−2t+2−2ln⁡t−1, the sign of g 4 ′ g 4 ′ g_(4)^(')g_{4}^{\prime}g4′ is given by t 2 − t − 2 t 2 − 2 t + 2 t 2 − t − 2 t 2 − 2 t + 2 t^(2)-t-2sqrt(t^(2)-2t+2)t^{2}-t-2 \sqrt{t^{2}-2 t+2}t2−t−2t2−2t+2; but ( t 2 − t ) 2 − 4 ( t 2 − 2 t + 2 ) = ( t − 10 ) 4 + 38 ( t − 10 ) 3 + 537 ( t − 10 ) 2 + 3348 ( t − 10 ) + 7772 > 0 t 2 − t 2 − 4 t 2 − 2 t + 2 = ( t − 10 ) 4 + 38 ( t − 10 ) 3 + 537 ( t − 10 ) 2 + 3348 ( t − 10 ) + 7772 > 0 (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(t2−t)2−4(t2−2t+2)=(t−10)4+38(t−10)3+537(t−10)2+3348(t−10)+7772>0 for t ≥ 10 t ≥ 10 t >= 10t \geq 10t≥10. It follows that g 3 ( t ) ≥ g 3 ( 10 ) = 3.45 … > 0 g 3 ( t ) ≥ g 3 ( 10 ) = 3.45 … > 0 g_(3)(t) >= g_(3)(10)=3.45 dots > 0g_{3}(t) \geq g_{3}(10)=3.45 \ldots>0g3(t)≥g3(10)=3.45…>0, hence g 1 g 1 g_(1)g_{1}g1 is positive for t ≥ 10 t ≥ 10 t >= 10t \geq 10t≥10 too.
In conclusion, g ( t ) > g ( 1 ) = 0 g ( t ) > g ( 1 ) = 0 g(t) > g(1)=0g(t)>g(1)=0g(t)>g(1)=0, and f 42 ( t ) > 1 / 2 f 42 ( t ) > 1 / 2 f_(42)(t) > 1//2f_{42}(t)>1 / 2f42(t)>1/2 for t > 1 t > 1 t > 1t>1t>1; in addition, lim t → 1 f 42 ( t ) = 1 / 2 lim t → 1   f 42 ( t ) = 1 / 2 lim_(t rarr1)f_(42)(t)=1//2\lim _{t \rightarrow 1} f_{42}(t)=1 / 2limt→1f42(t)=1/2.

3. ANOTHER INEQUALITY

In general it is not an easy task to find the range where the parameter α α alpha\alphaα 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 α ∈ ( 0 , 1 ) α ∈ ( 0 , 1 ) alpha in(0,1)\alpha \in(0,1)α∈(0,1) for which
(13) Q ( t ) − α A ( t ) − ( 1 − α ) S ( t ) > 0 , ∀ t > 1 . (13) Q ( t ) − α A ( t ) − ( 1 − α ) S ( t ) > 0 , ∀ t > 1 . {:(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*}(13)Q(t)−αA(t)−(1−α)S(t)>0,∀t>1.
In order to find the minimal value of α α alpha\alphaα for which (13) holds, we shall develop asymptotically the function F ( t , α ) = Q ( t ) − α A ( t ) − ( 1 − α ) S ( t ) F ( t , α ) = Q ( t ) − α A ( t ) − ( 1 − α ) S ( t ) F(t,alpha)=Q(t)-alpha A(t)-(1-alpha)S(t)F(t, \alpha)=Q(t)-\alpha A(t)-(1-\alpha) S(t)F(t,α)=Q(t)−αA(t)−(1−α)S(t) for t → ∞ t → ∞ t rarr oot \rightarrow \inftyt→∞. We denote
(14) u = F ( t , α ) = 1 2 2 + 2 t 2 − 1 2 α ( 1 + t ) − ( 1 − α ) t t 1 + t (14) u = F ( t , α ) = 1 2 2 + 2 t 2 − 1 2 α ( 1 + t ) − ( 1 − α ) t t 1 + t {:(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*}(14)u=F(t,α)=122+2t2−12α(1+t)−(1−α)tt1+t
Using the command asympt ( u , t , 3 u , t , 3 u,t,3u, t, 3u,t,3 ) we obtain the series
(15) ( 2 2 + α 2 − 1 ) t − α 2 + ( 1 − α ) ln ⁡ ( t ) + 2 / 4 − ( 1 − α ) ( ln ⁡ ( t ) + 1 / 2 ( ln ⁡ ( t ) ) 2 ) t + O ( t − 2 ) (15) 2 2 + α 2 − 1 t − α 2 + ( 1 − α ) ln ⁡ ( t ) + 2 / 4 − ( 1 − α ) ln ⁡ ( t ) + 1 / 2 ( ln ⁡ ( t ) ) 2 t + O t − 2 {:[(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*}(15)(22+α2−1)t−α2+(1−α)ln⁡(t)+2/4−(1−α)(ln⁡(t)+1/2(ln⁡(t))2)t+O(t−2)
(We mention that the term O ( t − 2 ) O t − 2 O(t^(-2))\mathcal{O}\left(t^{-2}\right)O(t−2) is used in the sense of Maple; from the point of view of Landau notation it should be O ( t − 2 ln 3 ⁡ ( t ) ) O t − 2 ln 3 ⁡ ( t ) O(t^(-2)ln^(3)(t))\mathcal{O}\left(t^{-2} \ln ^{3}(t)\right)O(t−2ln3⁡(t)).)
For u > 0 u > 0 u > 0u>0u>0, the condition 2 / 2 + α / 2 − 1 ≥ 0 2 / 2 + α / 2 − 1 ≥ 0 sqrt2//2+alpha//2-1 >= 0\sqrt{2} / 2+\alpha / 2-1 \geq 02/2+α/2−1≥0, hence α ≥ 2 − 2 α ≥ 2 − 2 alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2}α≥2−2, is obviously necessary. Now that we know the expected minimal value of α α alpha\alphaα, we state
Theorem 5. The inequality (13) holds if and only if α ≥ 2 − 2 α ≥ 2 − 2 alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2}α≥2−2.
Proof. We shall prove that the inequality holds for α = 2 − 2 α = 2 − 2 alpha=2-sqrt2\alpha=2-\sqrt{2}α=2−2 (hence a a aaa fortiori for α ≥ 2 − 2 α ≥ 2 − 2 alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2}α≥2−2 ).
Let us denote f ( t ) = ( 2 + 1 ) F ( t , 2 − 2 ) f ( t ) = ( 2 + 1 ) F ( t , 2 − 2 ) f(t)=(sqrt2+1)F(t,2-sqrt2)f(t)=(\sqrt{2}+1) F(t, 2-\sqrt{2})f(t)=(2+1)F(t,2−2), where F F FFF is given in (14). We have to prove that f ( t ) > 0 f ( t ) > 0 f(t) > 0f(t)>0f(t)>0 for t > 1 t > 1 t > 1t>1t>1. It follows that
f ( t ) = ( 2 + 1 ) 2 ( t 2 + 1 ) 2 − 2 ( t + 1 ) 2 − t t t + 1 f ( t ) = ( 2 + 1 ) 2 t 2 + 1 2 − 2 ( t + 1 ) 2 − t t t + 1 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}}f(t)=(2+1)2(t2+1)2−2(t+1)2−ttt+1
We put in the inequality ( 1 + x ) q < 1 + q x ( 1 + x ) q < 1 + q x (1+x)^(q) < 1+qx(1+x)^{q}<1+q x(1+x)q<1+qx, which holds for x > 0 , 0 < q < 1 x > 0 , 0 < q < 1 x > 0,0 < q < 1x>0,0<q<1x>0,0<q<1, x = t − 1 x = t − 1 x=t-1x=t-1x=t−1 and q = t / ( t + 1 ) q = t / ( t + 1 ) q=t//(t+1)q=t /(t+1)q=t/(t+1). It follows that
t t t + 1 < t 2 + 1 t + 1 t t t + 1 < t 2 + 1 t + 1 t^((t)/(t+1)) < (t^(2)+1)/(t+1)t^{\frac{t}{t+1}}<\frac{t^{2}+1}{t+1}ttt+1<t2+1t+1
and
f ( t ) > ( 1 + 2 ) 2 ( t + 1 ) ( ( t + 1 ) 2 + 2 t 2 − 2 ( t 2 + 2 ( 2 − 1 ) t + 1 ) ) f ( t ) > ( 1 + 2 ) 2 ( t + 1 ) ( t + 1 ) 2 + 2 t 2 − 2 t 2 + 2 ( 2 − 1 ) t + 1 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)f(t)>(1+2)2(t+1)((t+1)2+2t2−2(t2+2(2−1)t+1))
Let us denote the positive expressions
f 1 ( t ) = ( t + 1 ) 2 + 2 t 2 , f 2 ( t ) = 2 ( t 2 + 2 ( 2 − 1 ) t + 1 ) ; f 1 ( t ) = ( t + 1 ) 2 + 2 t 2 , f 2 ( t ) = 2 t 2 + 2 ( 2 − 1 ) t + 1 ; 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) ;f1(t)=(t+1)2+2t2,f2(t)=2(t2+2(2−1)t+1);
it follows easily that f 1 2 ( t ) − f 2 2 ( t ) = 4 t ( t − 1 ) 2 f 1 2 ( t ) − f 2 2 ( t ) = 4 t ( t − 1 ) 2 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}f12(t)−f22(t)=4t(t−1)2, therefore f ( t ) > 0 f ( t ) > 0 f(t) > 0f(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.
Received by the editors: January 12, 2006.
  1. *"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.
2006

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