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 \(\alpha, \beta \in R\) for which the bilateral inequalities \(\alpha H(a,b) + (1-\alpha)Q(a,b)
<
G(a,b)
<
\beta H(a,b) + (1-\beta)Q(a,b)\) hold \(\forall 0<\alpha<\beta\). 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.
About this paper
Journal
Studia Universitatis Babes-Bolyai, Mathematica
Publisher Name
Babes-Bolyai Univeristy Cluj-Napoca, Romania
Print ISSN
Online ISSN
2065-961x
google scholar link
[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.
Bilateral inequalities for harmonic, geometric and Hölder means
Abstract
For 0 < a < b0<a<b, the harmonic, geometric and Hölder means satisfy H < G < QH< G<Q. They are special cases ( p=-1,0,2p=-1,0,2 ) of power means M_(p)M_{p}. We consider the following problem: find all alpha,beta inR\alpha, \beta \in \mathbb{R} for which the bilateral inequalities 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) hold AA0 < a < b\forall 0<a<b. Then we replace in the bilateral inequalities the mean QQ by M_(p),p > 0M_{p}, p>0 and address the same problem.
Mathematics Subject Classification (2010): 26E60.
Keywords: Means, power means, bilateral inequalities.
Keywords: Means, power means, bilateral inequalities.
1. Introduction
We consider bivariate means m:R_(+)^(2)rarrRm: \mathbb{R}_{+}^{2} \rightarrow \mathbb{R} which are symmetric ( m(b,a)=m(a,b)m(b, a)= m(a, b) for all a,b > 0)a, b>0) and homogeneous (m(lambda a,lambda b)=lambda m(a,b)(m(\lambda a, \lambda b)=\lambda m(a, b) for all a,b,lambda > 0)a, b, \lambda>0).
For two means m_(1)m_{1} and m_(2)m_{2} we write m_(1) <= m_(2)m_{1} \leq m_{2} if and only if m_(1)(a,b) <= m_(2)(a,b)m_{1}(a, b) \leq m_{2}(a, b) for every a,b > 0a, b>0, and m_(1) < m_(2)m_{1}<m_{2} if and only if m_(1)(a,b) < m_(2)(a,b)m_{1}(a, b)<m_{2}(a, b) for all a,b > 0a, b>0 with a!=ba \neq b.
Since we are dealing with strict inequalities, we may and shall assume in the following that 0 < a < b0<a<b.
We consider the bivariate means
{:[(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*}
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
where min and max are the trivial means given by (a,b)|->min(a,b)(a, b) \mapsto \min (a, b) and (a,b)|->max(a,b)(a, b) \mapsto \max (a, b). The power means are monotonic in pp, and M_(-1)=H,M_(0)=G,M_(1)=AM_{-1}=H, M_{0}=G, M_{1}=A, and M_(2)=QM_{2}=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,AG, A and QQ.
Theorem 1.1. [2] The double inequality
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<b
holds if and only if alpha >= 1//2\alpha \geq 1 / 2 and beta <= 1-sqrt2//2\beta \leq 1-\sqrt{2} / 2.
In what follows we shall prove a similar result for the means H,GH, G and QQ. Afterwards we consider the more general case of the means H,GH, G and M_(p),p > 0M_{p}, p>0. We show that for p=5//2p=5 / 2 the auxiliary function ff is still monotone and we formulate an open problem.
In what follows we shall prove a similar result for the means H,GH, G and QQ. Afterwards we consider the more general case of the means H,GH, G and M_(p),p > 0M_{p}, p>0. We show that for p=5//2p=5 / 2 the auxiliary function ff is still monotone and we formulate an open problem.
2. Main result
For 0 < a < b0<a<b, the geometric, harmonic and Hölder means satisfy H < G < QH<G<Q. We shall find all the values of alpha\alpha and beta\beta in order that the geometric mean to be strictly between the combination of HH and QQ with parameters alpha\alpha, respectively beta\beta. Due to the homogeneity of all the means considered here, we may denote t=b//a,t > 1t=b / a, t>1, and write in the following m(t)m(t) instead of 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}, the double inequality
{:(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*}
is equivalent to
{:(2.2)beta < f(t) < alpha:}\begin{equation*}
\beta<f(t)<\alpha \tag{2.2}
\end{equation*}
where
{:(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*}
We shall prove the following result.
Theorem 2.1. The double inequality
Theorem 2.1. The double inequality
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>1
holds if and only if alpha >= 1\alpha \geq 1 and beta <= 2//3\beta \leq 2 / 3. The function
f_(1)(t)=(Q(t)-G(t))/(Q(t)-H(t))f_{1}(t)=\frac{Q(t)-G(t)}{Q(t)-H(t)}
is strictly increasing on ( 1,oo1, \infty ).
Proof. The function f_(1)f_{1} is given by
{:(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*}
We substitute t=s^(2),s > 1t=s^{2}, s>1 and get
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}}
The numerator of the derivative of this expression is
{:[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}
and the denominator is obviously positive. We shall prove that
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}
is positive for s > 1s>1, hence f_(1)f_{1} is strictly increasing. We write g_(1)(s)=0g_{1}(s)=0 as
{:(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*}
square both sides and get
(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 .
Denoting by 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+1 we get
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-2831,
which has only one change of sign. We apply Descartes' rule of signs for h_(1)(s+4)h_{1}(s+4) and we obtain that the polynomial h_(1)(s)h_{1}(s) has a single root greater than 4 . We denote by k_(1)(s)k_{1}(s) the 6th degree polynomial in the left hand side of (2.5) and get
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-2831,
which has only one change of sign. We apply Descartes' rule of signs for h_(1)(s+4)h_{1}(s+4) and we obtain that the polynomial h_(1)(s)h_{1}(s) has a single root greater than 4 . We denote by k_(1)(s)k_{1}(s) the 6th degree polynomial in the left hand side of (2.5) and get
{:(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*}
Then the polynomial (2.6) is positive on s > 4s>4, hence g_(1)(s)=0g_{1}(s)=0 has no solutions on s > 1s>1. It follows that f_(1)f_{1} is strictly increasing on ( 1,oo1, \infty ). Since lim_(t rarr1)f_(1)(t)=2//3\lim _{t \rightarrow 1} f_{1}(t)=2 / 3 and lim_(t rarr oo)f_(1)(t)=1\lim _{t \rightarrow \infty} f_{1}(t)=1, the theorem is proved.
We try to see if a similar result can be obtained by taking instead of M_(2)=QM_{2}=Q another power mean. For p=5//2p=5 / 2 we can prove
Theorem 2.2. The double inequality
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>1
holds if and only if alpha >= 1\alpha \geq 1 and beta <= 5//7\beta \leq 5 / 7. The function
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)}
is strictly increasing on ( 1,oo1, \infty ).
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)-(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*}
By substituting t=s^(2),s > 1t=s^{2}, s>1 we get
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}}
We differentiate the above function and obtain its numerator
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)
the denominator being positive. We denote
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}
and we write g_(2)(s)=0g_{2}(s)=0 as
{:(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*}
We apply the 5 th power to both sides of (2.8) and get h_(2)(s)=0h_{2}(s)=0, where
{:[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}
Using the Sturm sequence, we obtain that h_(2)(s)h_{2}(s) has no roots in (1,oo)(1, \infty). It follows that h_(2)(s) > 0h_{2}(s)>0 on ( 1,oo1, \infty ), and the derivative of f_(2)(t)f_{2}(t) is positive on this interval, hence f_(2)(t)f_{2}(t) is strictly increasing. Since 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)=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)=\frac{M_{p}(t)-G(t)}{M_{p}(t)-H(t)}
for arbitrary p > 0p>0. It is easy to check that lim_(t rarr1)f_(3)(t)=p//(p+1)\lim _{t \rightarrow 1} f_{3}(t)=p /(p+1) and lim_(t rarr oo)f_(3)(t)=\lim _{t \rightarrow \infty} f_{3}(t)= 1. It remains to study the monotonicity of f_(3)f_{3}. In the following theorem we prove that, for p > 5//2p>5 / 2, the function f_(3)f_{3} is not monotone on ( 1,oo1, \infty ).
Theorem 2.4. For p > 5//2p>5 / 2, the infimum of the function f_(3)f_{3} on ( 1,oo1, \infty ) satisfies the inequality
i n f_(t > 1)f_(3)(t) < (p)/(p+1)\inf _{t>1} f_{3}(t)<\frac{p}{p+1}
Proof. Let p > 5//2p>5 / 2. The function f_(3)f_{3} 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)-(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}}
and after the substitution t=s^(2),s > 1t=s^{2}, s>1 we get
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}}
The Taylor series for s_(0)=1s_{0}=1 reads
(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 1
and its derivative will be
-(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)
It follows that the derivative is negative at least for s > 1s>1 close to 1 , hence f_(3)f_{3} decreases and i n f_(t > 1)f_(3)(t) < p//(p+1)\inf _{t>1} f_{3}(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} is strictly increasing on ( 1,oo1, \infty ) for each p in(0,5//2]p \in(0,5 / 2]. Then, for each p in(0,5//2]p \in(0,5 / 2], the double inequality
Open problem. Prove that the function f_(3)f_{3} is strictly increasing on ( 1,oo1, \infty ) for each p in(0,5//2]p \in(0,5 / 2]. Then, for each p in(0,5//2]p \in(0,5 / 2], the double inequality
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>1
will be true if and only if alpha >= 1\alpha \geq 1 and beta <= p//(p+1)\beta \leq 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.
[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
"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
"Babeş-Bolyai" University
Faculty of Mathematics and Computer Sciences
1, Kogălniceanu Street, 400084 Cluj-Napoca, Romania
e-mail: anisiu@math.ubbcluj.ro
- This paper was presented at the 10th Joint Conference on Mathematics and Computer Science (MaCS 2014), May 21-25, 2014, Cluj-Napoca, Romania.