Bilateral inequalities for harmonic, geometric and Hölder means

Mira Anisiu
(original), paper

Abstract

For \(0<a<b\), the harmonic, geometric and Holder means satisfy \newline\( H<G<Q\). They are special cases \((p=-1,0,2)\) of power means \(M_{p}\).
We consider the following problem: find all \(\U{3b1} ,\U{3b2} \in R\) for which the bilateral inequalities  \(\U{3b1} H(a,b)+(1-\U{3b1} )Q(a,b)<G(a,b)<\U{3b2} H(a,b)+(1-\U{3b2} )Q(a,b)\) hold \(\forall0<a<b\). Then we replace in the bilateral inequalities the mean \(Q\) by \(Mp\), \(p>0\) and address the same problem

Authors

Mira-Cristiana Anisiu
“Tiberiu Popoviciu” Institute of Numerical Analysis Romanian Academy

Valeriu Anisiu
“Babes-Bolyai” University Faculty of Mathematics and Computer Sciences

Keywords

Means; power means; bilateral inequalities

Paper coordinates

M.-C. Anisiu, V. Anisiu, Bilateral inequalities for harmonic, geometric and Hölder means, Stud. Univ. Babes-Bolyai Math., 59 (2014) no. 4, pp. 463-468.

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Studia Universitatis Babes-Bolyai, Mathematica

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Babes-Bolyai Univeristy Cluj-Napoca, Romania

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https://www.cs.ubbcluj.ro/journal/studia-mathematica/archive/2014-4/06-Anisiu-final.pdf

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[1] Alzer, H., Qiu, S.-L., Inequalities for means in two variables, Arch. Math. (Basel), 80(2003), 201-215.
[2] Anisiu, M.-C., Anisiu, V., Bilateral inequalities for means, Revue d’Analyse Numerique et de Theorie de l’Approximation, 42(2)(2013), 85-93.
[3] Bullen, P.S., Handbook of Means and Their Inequalities, 2nd edn., Mathematics and Its Applications, Springer, Berlin, 1987.
[4] Chu, Y.-M., Wang, M.-K., Gong, W.-M., Two sharp double inequalities for Seiffert mean, J. Inequal. Appl. 2010, 2010: Article ID 44.
[5] Neuman, E., S´andor, J., Companion inequalities for certain bivariate means, Appl. Anal. Discr. Math., 3(2009), 46–51.
[6] Xia, W.-F., Chu, Y.-M., Optimal inequalities related to the logarithmic, identric, arithmetic and harmonic means, Revue d’Analyse Numerique et de Theorie de l’Approximation, 39(2)(2010), 176-183.

06-Anisiu-final

Bilateral inequalities for harmonic, geometric and Hölder means

Mira-Cristiana Anisiu and Valeriu Anisiu

Abstract

For 0 < a < b 0 < a < b 0 < a < b0<a<b0<a<b, the harmonic, geometric and Hölder means satisfy H < G < Q H < G < Q H < G < QH< G<QH<G<Q. They are special cases ( p = − 1 , 0 , 2 p = − 1 , 0 , 2 p=-1,0,2p=-1,0,2p=−1,0,2 ) of power means M p M p M_(p)M_{p}Mp. We consider the following problem: find all α , β ∈ R α , β ∈ R alpha,beta inR\alpha, \beta \in \mathbb{R}α,β∈R for which the bilateral inequalities α H ( a , b ) + ( 1 − α ) Q ( a , b ) < G ( a , b ) < β H ( a , b ) + ( 1 − β ) Q ( a , b ) α H ( a , b ) + ( 1 − α ) Q ( a , b ) < G ( a , b ) < β H ( a , b ) + ( 1 − β ) Q ( a , b ) alpha H(a,b)+(1-alpha)Q(a,b) < G(a,b) < beta H(a,b)+(1-beta)Q(a,b)\alpha H(a, b)+(1-\alpha) Q(a, b)<G(a, b)<\beta H(a, b)+(1-\beta) Q(a, b)αH(a,b)+(1−α)Q(a,b)<G(a,b)<βH(a,b)+(1−β)Q(a,b) hold ∀ 0 < a < b ∀ 0 < a < b AA0 < a < b\forall 0<a<b∀0<a<b. Then we replace in the bilateral inequalities the mean Q Q QQQ by M p , p > 0 M p , p > 0 M_(p),p > 0M_{p}, p>0Mp,p>0 and address the same problem.

Mathematics Subject Classification (2010): 26E60.
Keywords: Means, power means, bilateral inequalities.

1. Introduction

We consider bivariate means m : R + 2 → R m : R + 2 → R m:R_(+)^(2)rarrRm: \mathbb{R}_{+}^{2} \rightarrow \mathbb{R}m:R+2→R which are symmetric ( m ( b , a ) = m ( a , b ) m ( b , a ) = m ( a , b ) m(b,a)=m(a,b)m(b, a)= m(a, b)m(b,a)=m(a,b) for all a , b > 0 ) a , b > 0 ) a,b > 0)a, b>0)a,b>0) and homogeneous ( m ( λ a , λ b ) = λ m ( a , b ) ( m ( λ a , λ b ) = λ m ( a , b ) (m(lambda a,lambda b)=lambda m(a,b)(m(\lambda a, \lambda b)=\lambda m(a, b)(m(λa,λb)=λm(a,b) for all a , b , λ > 0 ) a , b , λ > 0 ) a,b,lambda > 0)a, b, \lambda>0)a,b,λ>0).
For two means m 1 m 1 m_(1)m_{1}m1 and m 2 m 2 m_(2)m_{2}m2 we write m 1 ≤ m 2 m 1 ≤ m 2 m_(1) <= m_(2)m_{1} \leq m_{2}m1≤m2 if and only if m 1 ( a , b ) ≤ m 2 ( a , b ) m 1 ( a , b ) ≤ m 2 ( a , b ) m_(1)(a,b) <= m_(2)(a,b)m_{1}(a, b) \leq m_{2}(a, b)m1(a,b)≤m2(a,b) for every a , b > 0 a , b > 0 a,b > 0a, b>0a,b>0, and m 1 < m 2 m 1 < m 2 m_(1) < m_(2)m_{1}<m_{2}m1<m2 if and only if m 1 ( a , b ) < m 2 ( a , b ) m 1 ( a , b ) < m 2 ( a , b ) m_(1)(a,b) < m_(2)(a,b)m_{1}(a, b)<m_{2}(a, b)m1(a,b)<m2(a,b) for all a , b > 0 a , b > 0 a,b > 0a, b>0a,b>0 with a ≠ b a ≠ b a!=ba \neq ba≠b.
Since we are dealing with strict inequalities, we may and shall assume in the following that 0 < a < b 0 < a < b 0 < a < b0<a<b0<a<b.
We consider the bivariate means
(1.1) A ( a , b ) = a + b 2 ; G ( a , b ) = a b ; H ( a , b ) = 2 a b a + b ; Q ( a , b ) = ( a 2 + b 2 2 ) 1 / 2 ; (1.2) M p ( a , b ) = { ( a p + b p 2 ) 1 / p , for p ≠ 0 a b , for p = 0 (1.1) A ( a , b ) = a + b 2 ; G ( a , b ) = a b ; H ( a , b ) = 2 a b a + b ; Q ( a , b ) = a 2 + b 2 2 1 / 2 ; (1.2) M p ( a , b ) = a p + b p 2 1 / p ,  for  p ≠ 0 a b ,  for  p = 0 {:[(1.1)A(a","b)=(a+b)/(2);quad G(a","b)=sqrt(ab);quad H(a","b)=(2ab)/(a+b);quad Q(a","b)=((a^(2)+b^(2))/(2))^(1//2);],[(1.2)M_(p)(a","b)={[((a^(p)+b^(p))/(2))^(1//p)","" for "p!=0],[sqrt(ab)","" for "p=0]:}]:}\begin{gather*} A(a, b)=\frac{a+b}{2} ; \quad G(a, b)=\sqrt{a b} ; \quad H(a, b)=\frac{2 a b}{a+b} ; \quad Q(a, b)=\left(\frac{a^{2}+b^{2}}{2}\right)^{1 / 2} ; \tag{1.1}\\ M_{p}(a, b)=\left\{\begin{array}{l} \left(\frac{a^{p}+b^{p}}{2}\right)^{1 / p}, \text { for } p \neq 0 \\ \sqrt{a b}, \text { for } p=0 \end{array}\right. \tag{1.2} \end{gather*}(1.1)A(a,b)=a+b2;G(a,b)=ab;H(a,b)=2aba+b;Q(a,b)=(a2+b22)1/2;(1.2)Mp(a,b)={(ap+bp2)1/p, for p≠0ab, for p=0
which are known as the arithmetic, geometric, harmonic, Hölder and power means, respectively. Properties and comparison of standard means can be found in [3].
The means from (1.1) are comparable:
min < H < G < A < Q < max min < H < G < A < Q < max min < H < G < A < Q < max\min <H<G<A<Q<\maxmin<H<G<A<Q<max
where min and max are the trivial means given by ( a , b ) ↦ min ( a , b ) ( a , b ) ↦ min ( a , b ) (a,b)|->min(a,b)(a, b) \mapsto \min (a, b)(a,b)↦min(a,b) and ( a , b ) ↦ max ( a , b ) ( a , b ) ↦ max ( a , b ) (a,b)|->max(a,b)(a, b) \mapsto \max (a, b)(a,b)↦max(a,b). The power means are monotonic in p p ppp, and M − 1 = H , M 0 = G , M 1 = A M − 1 = H , M 0 = G , M 1 = A M_(-1)=H,M_(0)=G,M_(1)=AM_{-1}=H, M_{0}=G, M_{1}=AM−1=H,M0=G,M1=A, and M 2 = Q M 2 = Q M_(2)=QM_{2}=QM2=Q.
Recently, many bilateral inequalities between means have been proved ([1], [2], [4], [5], [6]). We mention one of them, which was the starting point for this paper, and refers to the means G , A G , A G,AG, AG,A and Q Q QQQ.
Theorem 1.1. [2] The double inequality
α G ( a , b ) + ( 1 − α ) Q ( a , b ) < A ( a , b ) < β G ( a , b ) + ( 1 − β ) Q ( a , b ) , ∀ 0 < a < b α G ( a , b ) + ( 1 − α ) Q ( a , b ) < A ( a , b ) < β G ( a , b ) + ( 1 − β ) Q ( a , b ) , ∀ 0 < a < b alpha G(a,b)+(1-alpha)Q(a,b) < A(a,b) < beta G(a,b)+(1-beta)Q(a,b),AA0 < a < b\alpha G(a, b)+(1-\alpha) Q(a, b)<A(a, b)<\beta G(a, b)+(1-\beta) Q(a, b), \forall 0<a<bαG(a,b)+(1−α)Q(a,b)<A(a,b)<βG(a,b)+(1−β)Q(a,b),∀0<a<b
holds if and only if α ≥ 1 / 2 α ≥ 1 / 2 alpha >= 1//2\alpha \geq 1 / 2α≥1/2 and β ≤ 1 − 2 / 2 β ≤ 1 − 2 / 2 beta <= 1-sqrt2//2\beta \leq 1-\sqrt{2} / 2β≤1−2/2.
In what follows we shall prove a similar result for the means H , G H , G H,GH, GH,G and Q Q QQQ. Afterwards we consider the more general case of the means H , G H , G H,GH, GH,G and M p , p > 0 M p , p > 0 M_(p),p > 0M_{p}, p>0Mp,p>0. We show that for p = 5 / 2 p = 5 / 2 p=5//2p=5 / 2p=5/2 the auxiliary function f f fff is still monotone and we formulate an open problem.

2. Main result

For 0 < a < b 0 < a < b 0 < a < b0<a<b0<a<b, the geometric, harmonic and Hölder means satisfy H < G < Q H < G < Q H < G < QH<G<QH<G<Q. We shall find all the values of α α alpha\alphaα and β β beta\betaβ in order that the geometric mean to be strictly between the combination of H H HHH and Q Q QQQ with parameters α α alpha\alphaα, respectively β β beta\betaβ. Due to the homogeneity of all the means considered here, we may denote t = b / a , t > 1 t = b / a , t > 1 t=b//a,t > 1t=b / a, t>1t=b/a,t>1, and write in the following m ( t ) m ( t ) m(t)m(t)m(t) instead of m ( 1 , t ) = ( 1 / a ) m ( a , b ) m ( 1 , t ) = ( 1 / a ) m ( a , b ) m(1,t)=(1//a)m(a,b)m(1, t)=(1 / a) m(a, b)m(1,t)=(1/a)m(a,b). For any three means m 1 < m 2 < m 3 m 1 < m 2 < m 3 m_(1) < m_(2) < m_(3)m_{1}<m_{2}<m_{3}m1<m2<m3, the double inequality
(2.1) α m 1 ( t ) + ( 1 − α ) m 3 ( t ) < m 2 ( t ) < β m 1 ( t ) + ( 1 − β ) m 3 ( t ) (2.1) α m 1 ( t ) + ( 1 − α ) m 3 ( t ) < m 2 ( t ) < β m 1 ( t ) + ( 1 − β ) m 3 ( t ) {:(2.1)alpham_(1)(t)+(1-alpha)m_(3)(t) < m_(2)(t) < betam_(1)(t)+(1-beta)m_(3)(t):}\begin{equation*} \alpha m_{1}(t)+(1-\alpha) m_{3}(t)<m_{2}(t)<\beta m_{1}(t)+(1-\beta) m_{3}(t) \tag{2.1} \end{equation*}(2.1)αm1(t)+(1−α)m3(t)<m2(t)<βm1(t)+(1−β)m3(t)
is equivalent to
(2.2) β < f ( t ) < α (2.2) β < f ( t ) < α {:(2.2)beta < f(t) < alpha:}\begin{equation*} \beta<f(t)<\alpha \tag{2.2} \end{equation*}(2.2)β<f(t)<α
where
(2.3) f ( t ) = m 3 ( t ) − m 2 ( t ) m 3 ( t ) − m 1 ( t ) (2.3) f ( t ) = m 3 ( t ) − m 2 ( t ) m 3 ( t ) − m 1 ( t ) {:(2.3)f(t)=(m_(3)(t)-m_(2)(t))/(m_(3)(t)-m_(1)(t)):}\begin{equation*} f(t)=\frac{m_{3}(t)-m_{2}(t)}{m_{3}(t)-m_{1}(t)} \tag{2.3} \end{equation*}(2.3)f(t)=m3(t)−m2(t)m3(t)−m1(t)
We shall prove the following result.
Theorem 2.1. The double inequality
α H ( t ) + ( 1 − α ) Q ( t ) < G ( t ) < β H ( t ) + ( 1 − β ) Q ( t ) , ∀ t > 1 α H ( t ) + ( 1 − α ) Q ( t ) < G ( t ) < β H ( t ) + ( 1 − β ) Q ( t ) , ∀ t > 1 alpha H(t)+(1-alpha)Q(t) < G(t) < beta H(t)+(1-beta)Q(t),AA t > 1\alpha H(t)+(1-\alpha) Q(t)<G(t)<\beta H(t)+(1-\beta) Q(t), \forall t>1αH(t)+(1−α)Q(t)<G(t)<βH(t)+(1−β)Q(t),∀t>1
holds if and only if α ≥ 1 α ≥ 1 alpha >= 1\alpha \geq 1α≥1 and β ≤ 2 / 3 β ≤ 2 / 3 beta <= 2//3\beta \leq 2 / 3β≤2/3. The function
f 1 ( t ) = Q ( t ) − G ( t ) Q ( t ) − H ( t ) f 1 ( t ) = Q ( t ) − G ( t ) Q ( t ) − H ( t ) f_(1)(t)=(Q(t)-G(t))/(Q(t)-H(t))f_{1}(t)=\frac{Q(t)-G(t)}{Q(t)-H(t)}f1(t)=Q(t)−G(t)Q(t)−H(t)
is strictly increasing on ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ).
Proof. The function f 1 f 1 f_(1)f_{1}f1 is given by
(2.4) f 1 ( t ) = ( ( 2 t 2 + 2 ) 1 / 2 − 2 t 1 / 2 ) ( t + 1 ) ( 2 t 2 + 2 ) 1 / 2 t + ( 2 t 2 + 2 ) 1 / 2 − 4 t . (2.4) f 1 ( t ) = 2 t 2 + 2 1 / 2 − 2 t 1 / 2 ( t + 1 ) 2 t 2 + 2 1 / 2 t + 2 t 2 + 2 1 / 2 − 4 t . {:(2.4)f_(1)(t)=(((2t^(2)+2)^(1//2)-2t^(1//2))(t+1))/((2t^(2)+2)^(1//2)t+(2t^(2)+2)^(1//2)-4t).:}\begin{equation*} f_{1}(t)=\frac{\left(\left(2 t^{2}+2\right)^{1 / 2}-2 t^{1 / 2}\right)(t+1)}{\left(2 t^{2}+2\right)^{1 / 2} t+\left(2 t^{2}+2\right)^{1 / 2}-4 t} . \tag{2.4} \end{equation*}(2.4)f1(t)=((2t2+2)1/2−2t1/2)(t+1)(2t2+2)1/2t+(2t2+2)1/2−4t.
We substitute t = s 2 , s > 1 t = s 2 , s > 1 t=s^(2),s > 1t=s^{2}, s>1t=s2,s>1 and get
f 1 ( s 2 ) = ( ( 2 s 4 + 2 ) 1 / 2 − 2 s ) ( s 2 + 1 ) ( 2 s 4 + 2 ) 1 / 2 s 2 + ( 2 s 4 + 2 ) 1 / 2 − 4 s 2 f 1 s 2 = 2 s 4 + 2 1 / 2 − 2 s s 2 + 1 2 s 4 + 2 1 / 2 s 2 + 2 s 4 + 2 1 / 2 − 4 s 2 f_(1)(s^(2))=(((2s^(4)+2)^(1//2)-2s)(s^(2)+1))/((2s^(4)+2)^(1//2)s^(2)+(2s^(4)+2)^(1//2)-4s^(2))f_{1}\left(s^{2}\right)=\frac{\left(\left(2 s^{4}+2\right)^{1 / 2}-2 s\right)\left(s^{2}+1\right)}{\left(2 s^{4}+2\right)^{1 / 2} s^{2}+\left(2 s^{4}+2\right)^{1 / 2}-4 s^{2}}f1(s2)=((2s4+2)1/2−2s)(s2+1)(2s4+2)1/2s2+(2s4+2)1/2−4s2
The numerator of the derivative of this expression is
4 ( s 8 − 4 s 7 + 2 s 6 + 2 ( 2 s 4 + 2 ) 1 / 2 s 4 − 2 ( 2 s 4 + 2 ) 1 / 2 s 2 − 2 s 2 + 4 s − 1 ) = 4 ( s 2 − 1 ) ( s 6 − 4 s 5 + 3 s 4 − 4 s 3 + 3 s 2 − 4 s + 1 + 2 ( 2 s 4 + 2 ) 1 / 2 s 2 ) 4 s 8 − 4 s 7 + 2 s 6 + 2 2 s 4 + 2 1 / 2 s 4 − 2 2 s 4 + 2 1 / 2 s 2 − 2 s 2 + 4 s − 1 = 4 s 2 − 1 s 6 − 4 s 5 + 3 s 4 − 4 s 3 + 3 s 2 − 4 s + 1 + 2 2 s 4 + 2 1 / 2 s 2 {:[4(s^(8)-4s^(7)+2s^(6)+2(2s^(4)+2)^(1//2)s^(4)-2(2s^(4)+2)^(1//2)s^(2)-2s^(2)+4s-1)],[=4(s^(2)-1)(s^(6)-4s^(5)+3s^(4)-4s^(3)+3s^(2)-4s+1+2(2s^(4)+2)^(1//2)s^(2))]:}\begin{aligned} & 4\left(s^{8}-4 s^{7}+2 s^{6}+2\left(2 s^{4}+2\right)^{1 / 2} s^{4}-2\left(2 s^{4}+2\right)^{1 / 2} s^{2}-2 s^{2}+4 s-1\right) \\ & =4\left(s^{2}-1\right)\left(s^{6}-4 s^{5}+3 s^{4}-4 s^{3}+3 s^{2}-4 s+1+2\left(2 s^{4}+2\right)^{1 / 2} s^{2}\right) \end{aligned}4(s8−4s7+2s6+2(2s4+2)1/2s4−2(2s4+2)1/2s2−2s2+4s−1)=4(s2−1)(s6−4s5+3s4−4s3+3s2−4s+1+2(2s4+2)1/2s2)
and the denominator is obviously positive. We shall prove that
g 1 ( s ) = s 6 − 4 s 5 + 3 s 4 − 4 s 3 + 3 s 2 − 4 s + 1 + 2 ( 2 s 4 + 2 ) 1 / 2 s 2 g 1 ( s ) = s 6 − 4 s 5 + 3 s 4 − 4 s 3 + 3 s 2 − 4 s + 1 + 2 2 s 4 + 2 1 / 2 s 2 g_(1)(s)=s^(6)-4s^(5)+3s^(4)-4s^(3)+3s^(2)-4s+1+2(2s^(4)+2)^(1//2)s^(2)g_{1}(s)=s^{6}-4 s^{5}+3 s^{4}-4 s^{3}+3 s^{2}-4 s+1+2\left(2 s^{4}+2\right)^{1 / 2} s^{2}g1(s)=s6−4s5+3s4−4s3+3s2−4s+1+2(2s4+2)1/2s2
is positive for s > 1 s > 1 s > 1s>1s>1, hence f 1 f 1 f_(1)f_{1}f1 is strictly increasing. We write g 1 ( s ) = 0 g 1 ( s ) = 0 g_(1)(s)=0g_{1}(s)=0g1(s)=0 as
(2.5) s 6 − 4 s 5 + 3 s 4 − 4 s 3 + 3 s 2 − 4 s + 1 = − 2 ( 2 s 4 + 2 ) 1 / 2 s 2 , (2.5) s 6 − 4 s 5 + 3 s 4 − 4 s 3 + 3 s 2 − 4 s + 1 = − 2 2 s 4 + 2 1 / 2 s 2 , {:(2.5)s^(6)-4s^(5)+3s^(4)-4s^(3)+3s^(2)-4s+1=-2(2s^(4)+2)^(1//2)s^(2)",":}\begin{equation*} s^{6}-4 s^{5}+3 s^{4}-4 s^{3}+3 s^{2}-4 s+1=-2\left(2 s^{4}+2\right)^{1 / 2} s^{2}, \tag{2.5} \end{equation*}(2.5)s6−4s5+3s4−4s3+3s2−4s+1=−2(2s4+2)1/2s2,
square both sides and get
( s 8 − 4 s 7 − 4 s 5 + 6 s 4 − 4 s 3 − 4 s + 1 ) ( s − 1 ) 4 = 0 . s 8 − 4 s 7 − 4 s 5 + 6 s 4 − 4 s 3 − 4 s + 1 ( s − 1 ) 4 = 0 . (s^(8)-4s^(7)-4s^(5)+6s^(4)-4s^(3)-4s+1)(s-1)^(4)=0.\left(s^{8}-4 s^{7}-4 s^{5}+6 s^{4}-4 s^{3}-4 s+1\right)(s-1)^{4}=0 .(s8−4s7−4s5+6s4−4s3−4s+1)(s−1)4=0.
Denoting by h 1 ( s ) = s 8 − 4 s 7 − 4 s 5 + 6 s 4 − 4 s 3 − 4 s + 1 h 1 ( s ) = s 8 − 4 s 7 − 4 s 5 + 6 s 4 − 4 s 3 − 4 s + 1 h_(1)(s)=s^(8)-4s^(7)-4s^(5)+6s^(4)-4s^(3)-4s+1h_{1}(s)=s^{8}-4 s^{7}-4 s^{5}+6 s^{4}-4 s^{3}-4 s+1h1(s)=s8−4s7−4s5+6s4−4s3−4s+1 we get
h 1 ( s + 4 ) = s 8 + 28 s 7 + 336 s 6 + 2236 s 5 + 8886 s 4 + 20956 s 3 + 26640 s 2 + 12604 s − 2831 h 1 ( s + 4 ) = s 8 + 28 s 7 + 336 s 6 + 2236 s 5 + 8886 s 4 + 20956 s 3 + 26640 s 2 + 12604 s − 2831 h_(1)(s+4)=s^(8)+28s^(7)+336s^(6)+2236s^(5)+8886s^(4)+20956s^(3)+26640s^(2)+12604 s-2831h_{1}(s+4)=s^{8}+28 s^{7}+336 s^{6}+2236 s^{5}+8886 s^{4}+20956 s^{3}+26640 s^{2}+12604 s-2831h1(s+4)=s8+28s7+336s6+2236s5+8886s4+20956s3+26640s2+12604s−2831,
which has only one change of sign. We apply Descartes' rule of signs for h 1 ( s + 4 ) h 1 ( s + 4 ) h_(1)(s+4)h_{1}(s+4)h1(s+4) and we obtain that the polynomial h 1 ( s ) h 1 ( s ) h_(1)(s)h_{1}(s)h1(s) has a single root greater than 4 . We denote by k 1 ( s ) k 1 ( s ) k_(1)(s)k_{1}(s)k1(s) the 6th degree polynomial in the left hand side of (2.5) and get
(2.6) k 1 ( s + 4 ) = s 6 + 20 s 5 + 163 s 4 + 684 s 3 + 1523 s 2 + 1620 s + 545 . (2.6) k 1 ( s + 4 ) = s 6 + 20 s 5 + 163 s 4 + 684 s 3 + 1523 s 2 + 1620 s + 545 . {:(2.6)k_(1)(s+4)=s^(6)+20s^(5)+163s^(4)+684s^(3)+1523s^(2)+1620 s+545.:}\begin{equation*} k_{1}(s+4)=s^{6}+20 s^{5}+163 s^{4}+684 s^{3}+1523 s^{2}+1620 s+545 . \tag{2.6} \end{equation*}(2.6)k1(s+4)=s6+20s5+163s4+684s3+1523s2+1620s+545.
Then the polynomial (2.6) is positive on s > 4 s > 4 s > 4s>4s>4, hence g 1 ( s ) = 0 g 1 ( s ) = 0 g_(1)(s)=0g_{1}(s)=0g1(s)=0 has no solutions on s > 1 s > 1 s > 1s>1s>1. It follows that f 1 f 1 f_(1)f_{1}f1 is strictly increasing on ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ). Since lim t → 1 f 1 ( t ) = 2 / 3 lim t → 1   f 1 ( t ) = 2 / 3 lim_(t rarr1)f_(1)(t)=2//3\lim _{t \rightarrow 1} f_{1}(t)=2 / 3limt→1f1(t)=2/3 and lim t → ∞ f 1 ( t ) = 1 lim t → ∞   f 1 ( t ) = 1 lim_(t rarr oo)f_(1)(t)=1\lim _{t \rightarrow \infty} f_{1}(t)=1limt→∞f1(t)=1, the theorem is proved.
We try to see if a similar result can be obtained by taking instead of M 2 = Q M 2 = Q M_(2)=QM_{2}=QM2=Q another power mean. For p = 5 / 2 p = 5 / 2 p=5//2p=5 / 2p=5/2 we can prove
Theorem 2.2. The double inequality
α H ( t ) + ( 1 − α ) M 5 / 2 ( t ) < G ( t ) < β H ( t ) + ( 1 − β ) M 5 / 2 ( t ) , ∀ t > 1 α H ( t ) + ( 1 − α ) M 5 / 2 ( t ) < G ( t ) < β H ( t ) + ( 1 − β ) M 5 / 2 ( t ) , ∀ t > 1 alpha H(t)+(1-alpha)M_(5//2)(t) < G(t) < beta H(t)+(1-beta)M_(5//2)(t),quad AA t > 1\alpha H(t)+(1-\alpha) M_{5 / 2}(t)<G(t)<\beta H(t)+(1-\beta) M_{5 / 2}(t), \quad \forall t>1αH(t)+(1−α)M5/2(t)<G(t)<βH(t)+(1−β)M5/2(t),∀t>1
holds if and only if α ≥ 1 α ≥ 1 alpha >= 1\alpha \geq 1α≥1 and β ≤ 5 / 7 β ≤ 5 / 7 beta <= 5//7\beta \leq 5 / 7β≤5/7. The function
f 2 ( t ) = M 5 / 2 ( t ) − G ( t ) M 5 / 2 ( t ) − H ( t ) f 2 ( t ) = M 5 / 2 ( t ) − G ( t ) M 5 / 2 ( t ) − H ( t ) f_(2)(t)=(M_(5//2)(t)-G(t))/(M_(5//2)(t)-H(t))f_{2}(t)=\frac{M_{5 / 2}(t)-G(t)}{M_{5 / 2}(t)-H(t)}f2(t)=M5/2(t)−G(t)M5/2(t)−H(t)
is strictly increasing on ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ).
Proof. We have
(2.7) f 2 ( t ) = ( 1 2 t 5 / 2 + 1 2 ) 2 / 5 − t 1 / 2 ( 1 2 t 5 / 2 + 1 2 ) 2 / 5 − 2 t t + 1 (2.7) f 2 ( t ) = 1 2 t 5 / 2 + 1 2 2 / 5 − t 1 / 2 1 2 t 5 / 2 + 1 2 2 / 5 − 2 t t + 1 {:(2.7)f_(2)(t)=(((1)/(2)t^(5//2)+(1)/(2))^(2//5)-t^(1//2))/(((1)/(2)t^(5//2)+(1)/(2))^(2//5)-(2t)/(t+1)):}\begin{equation*} f_{2}(t)=\frac{\left(\frac{1}{2} t^{5 / 2}+\frac{1}{2}\right)^{2 / 5}-t^{1 / 2}}{\left(\frac{1}{2} t^{5 / 2}+\frac{1}{2}\right)^{2 / 5}-\frac{2 t}{t+1}} \tag{2.7} \end{equation*}(2.7)f2(t)=(12t5/2+12)2/5−t1/2(12t5/2+12)2/5−2tt+1
By substituting t = s 2 , s > 1 t = s 2 , s > 1 t=s^(2),s > 1t=s^{2}, s>1t=s2,s>1 we get
f 2 ( s 2 ) = ( ( 16 s 5 + 16 ) 2 / 5 − 4 s ) ( s 2 + 1 ) ( s 2 + 1 ) ( 16 s 5 + 16 ) 2 / 5 − 8 s 2 f 2 s 2 = 16 s 5 + 16 2 / 5 − 4 s s 2 + 1 s 2 + 1 16 s 5 + 16 2 / 5 − 8 s 2 f_(2)(s^(2))=(((16s^(5)+16)^(2//5)-4s)(s^(2)+1))/((s^(2)+1)(16s^(5)+16)^(2//5)-8s^(2))f_{2}\left(s^{2}\right)=\frac{\left(\left(16 s^{5}+16\right)^{2 / 5}-4 s\right)\left(s^{2}+1\right)}{\left(s^{2}+1\right)\left(16 s^{5}+16\right)^{2 / 5}-8 s^{2}}f2(s2)=((16s5+16)2/5−4s)(s2+1)(s2+1)(16s5+16)2/5−8s2
We differentiate the above function and obtain its numerator
32 ( s − 1 ) ( 2 s 8 − 6 s 7 − 2 s 6 − 2 s 5 − 2 s 3 − 2 s 2 − 6 s + 2 + s 2 ( s + 1 ) ( 16 s 5 + 16 ) 2 / 5 ) 32 ( s − 1 ) 2 s 8 − 6 s 7 − 2 s 6 − 2 s 5 − 2 s 3 − 2 s 2 − 6 s + 2 + s 2 ( s + 1 ) 16 s 5 + 16 2 / 5 32(s-1)(2s^(8)-6s^(7)-2s^(6)-2s^(5)-2s^(3)-2s^(2)-6s+2+s^(2)(s+1)(16s^(5)+16)^(2//5))32(s-1)\left(2 s^{8}-6 s^{7}-2 s^{6}-2 s^{5}-2 s^{3}-2 s^{2}-6 s+2+s^{2}(s+1)\left(16 s^{5}+16\right)^{2 / 5}\right)32(s−1)(2s8−6s7−2s6−2s5−2s3−2s2−6s+2+s2(s+1)(16s5+16)2/5)
the denominator being positive. We denote
g 2 ( s ) = 2 s 8 − 6 s 7 − 2 s 6 − 2 s 5 − 2 s 3 − 2 s 2 − 6 s + 2 + s 2 ( s + 1 ) ( 16 s 5 + 16 ) 2 / 5 g 2 ( s ) = 2 s 8 − 6 s 7 − 2 s 6 − 2 s 5 − 2 s 3 − 2 s 2 − 6 s + 2 + s 2 ( s + 1 ) 16 s 5 + 16 2 / 5 g_(2)(s)=2s^(8)-6s^(7)-2s^(6)-2s^(5)-2s^(3)-2s^(2)-6s+2+s^(2)(s+1)(16s^(5)+16)^(2//5)g_{2}(s)=2 s^{8}-6 s^{7}-2 s^{6}-2 s^{5}-2 s^{3}-2 s^{2}-6 s+2+s^{2}(s+1)\left(16 s^{5}+16\right)^{2 / 5}g2(s)=2s8−6s7−2s6−2s5−2s3−2s2−6s+2+s2(s+1)(16s5+16)2/5
and we write g 2 ( s ) = 0 g 2 ( s ) = 0 g_(2)(s)=0g_{2}(s)=0g2(s)=0 as
(2.8) 2 ( s 8 − 3 s 7 − s 6 − s 5 − s 3 − s 2 − 3 s + 1 ) s 2 ( s + 1 ) = − ( 16 s 5 + 16 ) 2 / 5 . (2.8) 2 s 8 − 3 s 7 − s 6 − s 5 − s 3 − s 2 − 3 s + 1 s 2 ( s + 1 ) = − 16 s 5 + 16 2 / 5 . {:(2.8)(2(s^(8)-3s^(7)-s^(6)-s^(5)-s^(3)-s^(2)-3s+1))/(s^(2)(s+1))=-(16s^(5)+16)^(2//5).:}\begin{equation*} \frac{2\left(s^{8}-3 s^{7}-s^{6}-s^{5}-s^{3}-s^{2}-3 s+1\right)}{s^{2}(s+1)}=-\left(16 s^{5}+16\right)^{2 / 5} . \tag{2.8} \end{equation*}(2.8)2(s8−3s7−s6−s5−s3−s2−3s+1)s2(s+1)=−(16s5+16)2/5.
We apply the 5 th power to both sides of (2.8) and get h 2 ( s ) = 0 h 2 ( s ) = 0 h_(2)(s)=0h_{2}(s)=0h2(s)=0, where
h 2 ( s ) = s 30 − 10 s 29 + 25 s 28 + 20 s 27 − 50 s 26 − 196 s 25 − 150 s 24 + 320 s 23 + 1305 s 22 + 2090 s 21 + 2439 s 20 + 2320 s 19 + 2550 s 18 + 3460 s 17 + 4760 s 16 + 5240 s 15 + 4760 s 14 + 3460 s 13 + 2550 s 12 + 2320 s 11 + 2439 s 10 + 2090 s 9 + 1305 s 8 + 320 s 7 − 150 s 6 − 196 s 5 − 50 s 4 + 20 s 3 + 25 s 2 − 10 s + 1 h 2 ( s ) = s 30 − 10 s 29 + 25 s 28 + 20 s 27 − 50 s 26 − 196 s 25 − 150 s 24 + 320 s 23 + 1305 s 22 + 2090 s 21 + 2439 s 20 + 2320 s 19 + 2550 s 18 + 3460 s 17 + 4760 s 16 + 5240 s 15 + 4760 s 14 + 3460 s 13 + 2550 s 12 + 2320 s 11 + 2439 s 10 + 2090 s 9 + 1305 s 8 + 320 s 7 − 150 s 6 − 196 s 5 − 50 s 4 + 20 s 3 + 25 s 2 − 10 s + 1 {:[h_(2)(s)=s^(30)-10s^(29)+25s^(28)+20s^(27)-50s^(26)-196s^(25)-150s^(24)+320s^(23)],[+1305s^(22)+2090s^(21)+2439s^(20)+2320s^(19)+2550s^(18)+3460s^(17)+4760s^(16)],[+5240s^(15)+4760s^(14)+3460s^(13)+2550s^(12)+2320s^(11)+2439s^(10)+2090s^(9)],[+1305s^(8)+320s^(7)-150s^(6)-196s^(5)-50s^(4)+20s^(3)+25s^(2)-10 s+1]:}\begin{aligned} & h_{2}(s)=s^{30}-10 s^{29}+25 s^{28}+20 s^{27}-50 s^{26}-196 s^{25}-150 s^{24}+320 s^{23} \\ & +1305 s^{22}+2090 s^{21}+2439 s^{20}+2320 s^{19}+2550 s^{18}+3460 s^{17}+4760 s^{16} \\ & +5240 s^{15}+4760 s^{14}+3460 s^{13}+2550 s^{12}+2320 s^{11}+2439 s^{10}+2090 s^{9} \\ & +1305 s^{8}+320 s^{7}-150 s^{6}-196 s^{5}-50 s^{4}+20 s^{3}+25 s^{2}-10 s+1 \end{aligned}h2(s)=s30−10s29+25s28+20s27−50s26−196s25−150s24+320s23+1305s22+2090s21+2439s20+2320s19+2550s18+3460s17+4760s16+5240s15+4760s14+3460s13+2550s12+2320s11+2439s10+2090s9+1305s8+320s7−150s6−196s5−50s4+20s3+25s2−10s+1
Using the Sturm sequence, we obtain that h 2 ( s ) h 2 ( s ) h_(2)(s)h_{2}(s)h2(s) has no roots in ( 1 , ∞ ) ( 1 , ∞ ) (1,oo)(1, \infty)(1,∞). It follows that h 2 ( s ) > 0 h 2 ( s ) > 0 h_(2)(s) > 0h_{2}(s)>0h2(s)>0 on ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ), and the derivative of f 2 ( t ) f 2 ( t ) f_(2)(t)f_{2}(t)f2(t) is positive on this interval, hence f 2 ( t ) f 2 ( t ) f_(2)(t)f_{2}(t)f2(t) is strictly increasing. Since lim t → 1 f 2 ( t ) = 5 / 7 , lim t → ∞ f 2 ( t ) = 1 lim t → 1   f 2 ( t ) = 5 / 7 , lim t → ∞   f 2 ( t ) = 1 lim_(t rarr1)f_(2)(t)=5//7,lim_(t rarr oo)f_(2)(t)=1\lim _{t \rightarrow 1} f_{2}(t)=5 / 7, \lim _{t \rightarrow \infty} f_{2}(t)=1limt→1f2(t)=5/7,limt→∞f2(t)=1, the theorem is proved.
Remark 2.3. We can consider the function
f 3 ( t ) = M p ( t ) − G ( t ) M p ( t ) − H ( t ) f 3 ( t ) = M p ( t ) − G ( t ) M p ( t ) − H ( t ) f_(3)(t)=(M_(p)(t)-G(t))/(M_(p)(t)-H(t))f_{3}(t)=\frac{M_{p}(t)-G(t)}{M_{p}(t)-H(t)}f3(t)=Mp(t)−G(t)Mp(t)−H(t)
for arbitrary p > 0 p > 0 p > 0p>0p>0. It is easy to check that lim t → 1 f 3 ( t ) = p / ( p + 1 ) lim t → 1   f 3 ( t ) = p / ( p + 1 ) lim_(t rarr1)f_(3)(t)=p//(p+1)\lim _{t \rightarrow 1} f_{3}(t)=p /(p+1)limt→1f3(t)=p/(p+1) and lim t → ∞ f 3 ( t ) = lim t → ∞   f 3 ( t ) = lim_(t rarr oo)f_(3)(t)=\lim _{t \rightarrow \infty} f_{3}(t)=limt→∞f3(t)= 1. It remains to study the monotonicity of f 3 f 3 f_(3)f_{3}f3. In the following theorem we prove that, for p > 5 / 2 p > 5 / 2 p > 5//2p>5 / 2p>5/2, the function f 3 f 3 f_(3)f_{3}f3 is not monotone on ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ).
Theorem 2.4. For p > 5 / 2 p > 5 / 2 p > 5//2p>5 / 2p>5/2, the infimum of the function f 3 f 3 f_(3)f_{3}f3 on ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ) satisfies the inequality
inf t > 1 f 3 ( t ) < p p + 1 inf t > 1   f 3 ( t ) < p p + 1 i n f_(t > 1)f_(3)(t) < (p)/(p+1)\inf _{t>1} f_{3}(t)<\frac{p}{p+1}inft>1f3(t)<pp+1
Proof. Let p > 5 / 2 p > 5 / 2 p > 5//2p>5 / 2p>5/2. The function f 3 f 3 f_(3)f_{3}f3 is given by
f 3 ( t ) = ( 1 2 t p + 1 2 ) 1 / p − t 1 / 2 ( 1 2 t p + 1 2 ) 1 / p − 2 t t + 1 f 3 ( t ) = 1 2 t p + 1 2 1 / p − t 1 / 2 1 2 t p + 1 2 1 / p − 2 t t + 1 f_(3)(t)=(((1)/(2)t^(p)+(1)/(2))^(1//p)-t^(1//2))/(((1)/(2)t^(p)+(1)/(2))^(1//p)-(2t)/(t+1))f_{3}(t)=\frac{\left(\frac{1}{2} t^{p}+\frac{1}{2}\right)^{1 / p}-t^{1 / 2}}{\left(\frac{1}{2} t^{p}+\frac{1}{2}\right)^{1 / p}-\frac{2 t}{t+1}}f3(t)=(12tp+12)1/p−t1/2(12tp+12)1/p−2tt+1
and after the substitution t = s 2 , s > 1 t = s 2 , s > 1 t=s^(2),s > 1t=s^{2}, s>1t=s2,s>1 we get
f 3 ( s 2 ) = ( ( 1 2 s 2 p + 1 2 ) 1 / p − s ) ( s 2 + 1 ) ( s 2 + 1 ) ( 1 2 s 2 p + 1 2 ) 1 / p − 2 s 2 f 3 s 2 = 1 2 s 2 p + 1 2 1 / p − s s 2 + 1 s 2 + 1 1 2 s 2 p + 1 2 1 / p − 2 s 2 f_(3)(s^(2))=((((1)/(2)s^(2p)+(1)/(2))^(1//p)-s)(s^(2)+1))/((s^(2)+1)((1)/(2)s^(2p)+(1)/(2))^(1//p)-2s^(2))f_{3}\left(s^{2}\right)=\frac{\left(\left(\frac{1}{2} s^{2 p}+\frac{1}{2}\right)^{1 / p}-s\right)\left(s^{2}+1\right)}{\left(s^{2}+1\right)\left(\frac{1}{2} s^{2 p}+\frac{1}{2}\right)^{1 / p}-2 s^{2}}f3(s2)=((12s2p+12)1/p−s)(s2+1)(s2+1)(12s2p+12)1/p−2s2
The Taylor series for s 0 = 1 s 0 = 1 s_(0)=1s_{0}=1s0=1 reads
p p + 1 − p ( 2 p − 5 ) 12 ( p + 1 ) ( s − 1 ) 2 + p ( 2 p − 5 ) 12 ( p + 1 ) ( s − 1 ) 3 + O ( ( s − 1 ) 4 ) , for s → 1 p p + 1 − p ( 2 p − 5 ) 12 ( p + 1 ) ( s − 1 ) 2 + p ( 2 p − 5 ) 12 ( p + 1 ) ( s − 1 ) 3 + O ( s − 1 ) 4 ,  for  s → 1 (p)/(p+1)-(p(2p-5))/(12(p+1))(s-1)^(2)+(p(2p-5))/(12(p+1))(s-1)^(3)+O((s-1)^(4))," for "s rarr1\frac{p}{p+1}-\frac{p(2 p-5)}{12(p+1)}(s-1)^{2}+\frac{p(2 p-5)}{12(p+1)}(s-1)^{3}+O\left((s-1)^{4}\right), \text { for } s \rightarrow 1pp+1−p(2p−5)12(p+1)(s−1)2+p(2p−5)12(p+1)(s−1)3+O((s−1)4), for s→1
and its derivative will be
− p ( 2 p − 5 ) 6 ( p + 1 ) ( s − 1 ) + O ( ( s − 1 ) 2 ) − p ( 2 p − 5 ) 6 ( p + 1 ) ( s − 1 ) + O ( s − 1 ) 2 -(p(2p-5))/(6(p+1))(s-1)+O((s-1)^(2))-\frac{p(2 p-5)}{6(p+1)}(s-1)+O\left((s-1)^{2}\right)−p(2p−5)6(p+1)(s−1)+O((s−1)2)
It follows that the derivative is negative at least for s > 1 s > 1 s > 1s>1s>1 close to 1 , hence f 3 f 3 f_(3)f_{3}f3 decreases and inf t > 1 f 3 ( t ) < p / ( p + 1 ) inf t > 1   f 3 ( t ) < p / ( p + 1 ) i n f_(t > 1)f_(3)(t) < p//(p+1)\inf _{t>1} f_{3}(t)<p /(p+1)inft>1f3(t)<p/(p+1).
Based on the results in theorems 2.1 and 2.2, we formulate the following
Open problem. Prove that the function f 3 f 3 f_(3)f_{3}f3 is strictly increasing on ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ) for each p ∈ ( 0 , 5 / 2 ] p ∈ ( 0 , 5 / 2 ] p in(0,5//2]p \in(0,5 / 2]p∈(0,5/2]. Then, for each p ∈ ( 0 , 5 / 2 ] p ∈ ( 0 , 5 / 2 ] p in(0,5//2]p \in(0,5 / 2]p∈(0,5/2], the double inequality
α H ( t ) + ( 1 − α ) M p ( t ) < G ( t ) < β H ( t ) + ( 1 − β ) M p ( t ) , ∀ t > 1 α H ( t ) + ( 1 − α ) M p ( t ) < G ( t ) < β H ( t ) + ( 1 − β ) M p ( t ) , ∀ t > 1 alpha H(t)+(1-alpha)M_(p)(t) < G(t) < beta H(t)+(1-beta)M_(p)(t),AA t > 1\alpha H(t)+(1-\alpha) M_{p}(t)<G(t)<\beta H(t)+(1-\beta) M_{p}(t), \forall t>1αH(t)+(1−α)Mp(t)<G(t)<βH(t)+(1−β)Mp(t),∀t>1
will be true if and only if α ≥ 1 α ≥ 1 alpha >= 1\alpha \geq 1α≥1 and β ≤ p / ( p + 1 ) β ≤ p / ( p + 1 ) beta <= p//(p+1)\beta \leq p /(p+1)β≤p/(p+1).

References

[1] Alzer, H., Qiu, S.-L., Inequalities for means in two variables, Arch. Math. (Basel), 80(2003), 201-215.
[2] Anisiu, M.-C., Anisiu, V., Bilateral inequalities for means, Revue d'Analyse Numérique et de Théorie de l'Approximation, 42(2)(2013), 85-93.
[3] Bullen, P.S., Handbook of Means and Their Inequalities, 2nd edn., Mathematics and Its Applications, Springer, Berlin, 1987.
[4] Chu, Y.-M., Wang, M.-K., Gong, W.-M., Two sharp double inequalities for Seiffert mean, J. Inequal. Appl. 2010, 2010: Article ID 44.
[5] Neuman, E., Sándor, J., Companion inequalities for certain bivariate means, Appl. Anal. Discr. Math., 3(2009), 46-51.
[6] Xia, W.-F., Chu, Y.-M., Optimal inequalities related to the logarithmic, identric, arithmetic and harmonic means, Revue d'Analyse Numérique et de Théorie de l'Approximation, 39(2)(2010), 176-183.
Mira-Cristiana Anisiu
"Tiberiu Popoviciu" Institute of Numerical Analysis
Romanian Academy
P.O. Box 68, 400110 Cluj-Napoca, Romania
e-mail: mira@math.ubbcluj.ro
Valeriu Anisiu
"Babeş-Bolyai" University
Faculty of Mathematics and Computer Sciences
1, Kogălniceanu Street, 400084 Cluj-Napoca, Romania
e-mail: anisiu@math.ubbcluj.ro
  1. This paper was presented at the 10th Joint Conference on Mathematics and Computer Science (MaCS 2014), May 21-25, 2014, Cluj-Napoca, Romania.
2014

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