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Proof of an open inequality with double powerexponential functions
Journal of Inequalities and Applications volume 2013, Article number: 468 (2013)
Abstract
Cîrtoaje (J. Nonlinear Sci. Appl. 4(2):130137, 2011) conjectured that the inequality {a}^{{(2b)}^{x}}+{b}^{{(2a)}^{x}}\le 1 with double powerexponential functions holds for all nonnegative real numbers a, b with a+b=1 and all x\ge 1. In this paper, we shall prove the conjecture affirmatively.
MSC:26D10.
1 Introduction
The study of inequalities with powerexponential functions is one of the active areas of research in the mathematical analysis. Cîrtoaje [1, 2] studied inequalities with powerexponential functions and conjectured some open inequalities. He posed the open inequality as Conjecture 4.8 in [1],
which holds for all nonnegative real numbers a, b with a+b=1. He proved in [2] that this inequality holds. Moreover, he conjectured the more generalized inequality containing double powerexponential functions in [2]:
holds for all nonnegative real numbers a, b with a+b=1 and all x\ge 1, which is Conjecture 5.1 in [2] and still an open problem. Cîrtoaje’s open inequality (1.1) is an interesting and new problem of great importance in the power exponential inequality theory. In this paper, we shall prove the conjecture affirmatively. The following is our main theorem.
Theorem 1.1 For all nonnegative real numbers a, b with a+b=1 and all x\ge 1, inequality (1.1) holds.
We shall show this theorem by using differentiation mainly.
Let
and
Since F(x,0)=F(x,\frac{1}{2})=1 and F(x,a)=F(x,1a), it suffices to show that F(x,a)\le 1 for 0<a<\frac{1}{2} and x\ge 1. To prove Theorem 1.1, we shall investigate the sign of
We set
Then {F}_{x}(x,a) clearly has the same sign as G(x,a).
Since G(x,a) has the both signs, in order to get the sign of it, we need to investigate the signs of G(1,a) and {G}_{x}(1,a). We shall describe the results in Sections 2.1 and 2.2. Moreover, for fixed a\in (0,\frac{1}{2}), we show in Section 2.3 that {G}_{x}=\partial G/\partial x is strictly increasing for x\ge 1; that is, G(x,a) is convex on [1,\mathrm{\infty}), which is the main idea of our proof.
In Section 3, we consider the cases of {G}_{x}(1,a)\ge 0 and {G}_{x}(1,a)<0 to prove Theorem 1.1. Using three propositions given in Section 2, we notice the following (1) and (2).

(1)
From Proposition 2.8, for fixed a, if {G}_{x}(1,a)\ge 0, then {G}_{x}(x,a)>0 for x>1 and if {G}_{x}(1,a)<0, then there exists uniquely a number \tilde{x}>1 such that {G}_{x}(\tilde{x},a)=0.

(2)
From Propositions 2.5 and 2.6, we notice that G(1,a)<0 when {G}_{x}(1,a)<0.
(1) and (2) play an important role in the proof of Theorem 1.1.
We shall use the functions F(x,a) and G(x,a) defined here throughout this paper.
2 Preliminaries
2.1 The sign of G(1,a)
From the definition of G(x,a), we have
where
Then we have
In this subsection, we shall show that G(1,a)<0 for \frac{15}{100}\le a<\frac{1}{2}.
Consider first the case \frac{15}{100}\le a\le \frac{1}{4}. We have
and
where
and
Lemma 2.1 If \frac{15}{100}\le a\le \frac{1}{4}, then
Proof First, we show that
for \frac{15}{100}\le a\le \frac{1}{4}. Let {\lambda}_{1} and {\lambda}_{2} ({\lambda}_{1}<{\lambda}_{2}) be the solutions of
then we have
and
Since Q(a)\le 0 for {e}^{{\lambda}_{1}}\le a\le {e}^{{\lambda}_{2}}, we have Q(a)<0 for \frac{15}{100}\le a\le \frac{1}{4}. Since Q(b)\ge 0 for b\ge {e}^{{\lambda}_{2}}, that is, for a\le 1{e}^{{\lambda}_{2}}\cong 0.2505, we have Q(b)>0 for \frac{15}{100}\le a\le \frac{1}{4}. Therefore, from Q(a)<0 and Q(b)>0, we get {R}^{\u2033}(a)<0 for \frac{15}{100}\le a\le \frac{1}{4}.
Next, we show that
for \frac{15}{100}\le a\le \frac{1}{4}. If we set
then
and
Since {R}^{\u2033}(a)<0 for \frac{15}{100}\le a\le \frac{1}{4}, {f}^{\prime} is strictly decreasing on the interval [\frac{15}{100},\frac{1}{4}] and we have {f}^{\prime}(a)\ge {f}^{\prime}(\frac{1}{4})={R}^{\prime}(\frac{1}{4})5(\cong 0.0377)>0. Therefore, f is strictly increasing on the interval [\frac{15}{100},\frac{1}{4}], so we have f(a)\le f(\frac{1}{4})=R(\frac{1}{4})+1(\cong 0.0363)<0. Thus, we get R(a)<5(a\frac{1}{4})1 for \frac{15}{100}\le a\le \frac{1}{4}.
In order to complete the proof of this lemma, it suffices to show from the above inequality with respect to R(a) that g(a)<0 for \frac{15}{100}\le a\le \frac{1}{4}, where
We have
and
Since {g}^{\u2033}(a)>0, {g}^{\prime} is strictly increasing on the interval [\frac{15}{100},\frac{1}{4}]. Since {g}^{\prime}(\frac{15}{100})(\cong 2.9624)<0 and {g}^{\prime}(\frac{1}{4})(\cong 0.3187)>0, there exists uniquely a number c\in (\frac{15}{100},\frac{1}{4}) such that {g}^{\prime}(c)=0. Then we have {g}^{\prime}(a)<0 for \frac{15}{100}<a<c and {g}^{\prime}(a)>0 for c<a<\frac{1}{4}. Hence, g is strictly decreasing on the interval [\frac{15}{100},c] and strictly increasing on the interval [c,\frac{1}{4}]. Therefore, g(a)\le max\{g(\frac{15}{100}),g(\frac{1}{4})\}. Since g(\frac{15}{100})\cong 0.0582 and g(\frac{1}{4})\cong 0.1630, we can get g(a)<0 for \frac{15}{100}\le a\le \frac{1}{4}. □
It still remains to show that G(1,a)<0 for \frac{1}{4}\le a<\frac{1}{2}. Since
using the substitution
we need to prove that A(t)<0 for 0<t\le \frac{1}{2}, where
Lemma 2.2 If 0<t\le \frac{1}{2}, then
Proof We need to prove that f(t)>0 for 0<t\le \frac{1}{2}, where
We have
and
If we set g(t)={f}^{\u2034}(t)\times {(1+t)}^{3}{(1t)}^{3}, then we have
From
it follows that g is strictly decreasing on (0,\frac{1}{2}). Since g(0)=3 and g(\frac{1}{2})=\frac{23}{4}, there exists uniquely a number {c}_{1}\in (0,\frac{1}{2}) such that g({c}_{1})=0. Since g(t)>0 for 0<t<{c}_{1} and g(t)<0 for {c}_{1}<t<\frac{1}{2}, we have {f}^{\u2034}(t)>0 for 0<t<{c}_{1} and {f}^{\u2034}(t)<0 for {c}_{1}<t<\frac{1}{2}. It follows that {f}^{\u2033} is strictly increasing on the interval (0,{c}_{1}) and strictly decreasing on the interval ({c}_{1},\frac{1}{2}). Since {f}^{\u2033}(0)=0 and {f}^{\u2033}(\frac{1}{2})=2ln\frac{3}{2}\frac{19}{9}(\cong 1.3001)<0, there exists uniquely a number {c}_{2}\in (0,\frac{1}{2}) such that {f}^{\u2033}({c}_{2})=0. Since {f}^{\u2033}(t)>0 for 0<t<{c}_{2} and {f}^{\u2033}(t)<0 for {c}_{2}<t<\frac{1}{2}, {f}^{\prime} is strictly increasing on the interval (0,{c}_{2}) and strictly decreasing on the interval ({c}_{2},\frac{1}{2}). Since {f}^{\prime}(0)=0 and {f}^{\prime}(\frac{1}{2})=2ln\frac{3}{2}\frac{5}{6}(\cong 0.0224)<0, there exists uniquely a number {c}_{3}\in (0,\frac{1}{2}) such that {f}^{\prime}({c}_{3})=0. Hence, {f}^{\prime}(t)>0 for 0<t<{c}_{3} and {f}^{\prime}(t)<0 for {c}_{3}<t<\frac{1}{2}. Thus, f is strictly increasing on the interval (0,{c}_{3}) and strictly decreasing on the interval ({c}_{3},\frac{1}{2}). Since f(0)=0 and f(\frac{1}{2})=\frac{7}{4}ln\frac{3}{2}ln2\cong 0.0164, we can get f(t)>0 for 0<t\le \frac{1}{2}. □
Lemma 2.3 If 0<t\le \frac{1}{2}, then
Proof We need to show that f(t)<0, where
We have
and
where
and
We see that {(ln{S}_{2}(t))}^{\u2033} has the same sign as
where
and
We have
where
and
Since ln2>ln(1+t) and ln2>ln(1t) for 0<t\le \frac{1}{2}, we have {f}_{1}(t)>0. Since {f}_{2}^{\prime}(t)=2ln22ln(1{t}^{2})\le 4ln22ln3\cong 0.3260 for 0<t\le \frac{1}{2}, {f}_{2} is strictly decreasing on the interval (0,\frac{1}{2}]. Therefore, we can get {f}_{2}(t)<{lim}_{t\to 0}{f}_{2}(t)=0 for 0<t\le \frac{1}{2}. From {f}_{1}(t)>0 and {f}_{2}(t)<0, it follows that {B}_{1}(t)<0. Since {B}_{1}(t)<0, {B}_{2}(t)<0 and {B}_{3}(t)<0, we get B(t)<0, hence {(ln{S}_{2}(t))}^{\u2033}<0 for 0<t\le \frac{1}{2}. Thus, {f}^{\prime} is strictly decreasing on the interval (0,\frac{1}{2}] and we have {f}^{\prime}(t)<{lim}_{t\to 0}{f}^{\prime}(t)=0 for 0<t\le \frac{1}{2}. Since f is strictly decreasing on the interval (0,\frac{1}{2}], we have f(t)<{lim}_{t\to 0}f(t)=0 for 0<t\le \frac{1}{2}. Therefore, we get ln{S}_{2}(t)<\frac{2}{ln2}t. □
Lemma 2.4 If \frac{1}{4}\le a\le \frac{1}{2}, then
Proof We need to show that A(t)<0 for 0<t\le \frac{1}{2}. By Lemmas 2.2 and 2.3, it suffices to show that f(t)<0, where
We have
and
Since {f}^{\u2033}(t)>0 for 0<t\le \frac{1}{2}, {f}^{\prime} is strictly increasing on the interval (0,\frac{1}{2}]. Since {lim}_{t\to 0}{f}^{\prime}(t)=1+2ln2\frac{2}{ln2}(\cong 0.4990)<0 and {f}^{\prime}(\frac{1}{2})=4ln2+\frac{38}{21}ln3\frac{2}{ln2}(\cong 0.5981)>0, there exists uniquely a number c\in (0,\frac{1}{2}) such that {f}^{\prime}(c)=0. Hence, f is strictly decreasing on the interval (0,c) and strictly increasing on the interval (c,\frac{1}{2}]. Since {lim}_{t\to +0}f(t)=0 and f(\frac{1}{2})=\frac{1}{2}ln3+ln7\frac{1}{ln2}\cong 0.0460, we get f(t)<0. Therefore, A(t)<f(t)<0 for 0<t\le \frac{1}{2}. □
From Lemmas 2.1 and 2.4, we get the following result.
Proposition 2.5 If \frac{15}{100}\le a<\frac{1}{2}, then
We notice that {lim}_{a\to \frac{1}{2}0}G(1,a)=0.
2.2 The sign of {G}_{x}(1,a)
We have
hence
where b=1a.
Proposition 2.6 There exists a number c\in (\frac{15}{100},\frac{1}{2}) such that {G}_{x}(1,a)>0 for 0<a<c.
Proof Let us denote {G}_{x}(1,a) by f(a). We have
and
Since 0<a<\frac{1}{2}, we have blnbalna>0, ln(2a)<0, ln(2b)>0 and 12a>0. Therefore, if 2ln(2b)1\ge 0, then {f}^{\u2033}(a)>0. The condition 2ln(2b)1\ge 0 is true for 0<a<{a}_{0}, where
Consequently, {f}^{\prime} is strictly increasing on (0,{a}_{0}]. Since {f}^{\prime}({a}_{0})\cong 2.5412, it follows that {f}^{\prime}(a)<0 on (0,{a}_{0}], and f is strictly decreasing on (0,{a}_{0}]. Since {lim}_{a\to +0}f(a)=\mathrm{\infty} and f({a}_{0})\cong 0.0413, there exists uniquely a number c\in (0,{a}_{0}) such that f(c)=0. Then we have f(a)>0 for 0<a<c and f(a)<0 for c<a<{a}_{0}. Since \frac{15}{100}<{a}_{0} and f(\frac{15}{100})\cong 0.0354, we can get \frac{15}{100}<c. □
2.3 The convexity of G(x,a)
In order to investigate the convexity of G(x,a) with respect to x, we need the following lemma.
Lemma 2.7 If 0<a<\frac{1}{2}, then
where b=1a.
Proof We first show that the inequality
holds for 0<a<\frac{1}{2}. We denote
Then we have
We set
then we have
and
Therefore, {g}^{\prime} is strictly decreasing on the interval (0,\frac{1}{2}). Since {lim}_{a\to +0}{g}^{\prime}(a)=\mathrm{\infty} and {g}^{\prime}(\frac{1}{2})=2ln22<0, there exists uniquely a number c\in (0,\frac{1}{2}) such that {g}^{\prime}(c)=0. Since {g}^{\prime}(a)>0 for 0<a<c and {g}^{\prime}(a)<0 for c<a<\frac{1}{2}, g is strictly increasing on the interval (0,c) and strictly decreasing on the interval (c,\frac{1}{2}). Since {lim}_{a\to +0}g(a)=0 and g(\frac{1}{2})=0, we get g(a)>0 for 0<a<\frac{1}{2}. Therefore, we have {f}^{\prime}(a)>0 for any a\in (0,\frac{1}{2}). Since f is strictly increasing on the interval (0,\frac{1}{2}), we can get f(a)<f(\frac{1}{2})={(ln2)}^{2}. Hence, we have {(ln2)}^{2}>lnalnb for 0<a<\frac{1}{2}. Also, the inequality
is equivalent to
From lnblna>0 and {(ln2)}^{2}lnalnb>0, it follows that (lnblna)({(ln2)}^{2}lnalnb)>0. This completes the proof of the lemma. □
Proposition 2.8 If 0<a<\frac{1}{2} and x\ge 1, then {G}_{x} is strictly increasing with respect to x.
Proof For fixed a\in (0,\frac{1}{2}), let us denote f(x)={G}_{x}(x,a); that is,
Clearly, we need to show that {f}^{\prime}(x)>0 for x\ge 1. From
we can write the inequality {f}^{\prime}(x)>0 in the form
Since {(\frac{a}{b})}^{x}<1, it is enough to show that
which follows immediately from Lemma 2.7. □
3 Proof of Theorem 1.1
In this section, we shall prove the main theorem.
Proof We shall show that if 0\le a\le \frac{1}{2} and x\ge 1, then F(x,a)\le 1. The inequality F(1,a)\le 1 is proved in Cîrtoaje [2]. Since F(x,0)=F(x,\frac{1}{2})=1, we may show this inequality for 0<a<\frac{1}{2}. We note that {G}_{x}(1,a) has the both signs, so we consider the cases of {G}_{x}(1,a)\ge 0 and {G}_{x}(1,a)<0.

(1)
We first assume that {G}_{x}(1,a)\ge 0. We have {G}_{x}(x,a)>{G}_{x}(1,a) for x>1 from Proposition 2.8. Since we have {G}_{x}(x,a)>0 on the assumption, G is strictly increasing with respect to x. Therefore, G(x,a)>G(1,a) for x>1. We note that G(1,a) has the both signs, so we consider the cases of G(1,a)\ge 0 and G(1,a)<0.

(i)
If G(1,a)\ge 0, then G(x,a)>0, so we have {F}_{x}(x,a)>0 for x>1. Hence, F is strictly increasing with respect to x, and we have F(x,a)\le {lim}_{x\to \mathrm{\infty}}F(x,a)=1.

(ii)
If G(1,a)<0, then since {lim}_{x\to \mathrm{\infty}}G(x,a)=\mathrm{\infty}, there exists uniquely a number {x}_{1}>1 satisfying G({x}_{1},a)=0. Since G(x,a)<0 for 1<x<{x}_{1} and G(x,a)>0 for x>{x}_{1}, we have {F}_{x}(x,a)<0 for 1<x<{x}_{1} and {F}_{x}(x,a)>0 for x>{x}_{1}. Hence, F is strictly decreasing for 1<x<{x}_{1} and strictly increasing for x>{x}_{1}. Therefore, we get
F(x,a)\le max\{F(1,a),\underset{x\to \mathrm{\infty}}{lim}F(x,a)\}=1.

(i)

(2)
We next assume that {G}_{x}(1,a)<0. Since {G}_{x} is strictly increasing with respect to x from Proposition 2.8 and {lim}_{x\to \mathrm{\infty}}{G}_{x}(x,a)=\mathrm{\infty}, there exists uniquely a number {x}_{2}>1 satisfying {G}_{x}({x}_{2},a)=0. Since {G}_{x}(x,a)<0 for 1<x<{x}_{2} and {G}_{x}(x,a)>0 for x>{x}_{2}, G is strictly decreasing for 1<x<{x}_{2} and strictly increasing for x>{x}_{2}. By the assumption {G}_{x}(1,a)<0 and Proposition 2.6, it follows that a>\frac{15}{100}. Hence, G(1,a)<0 by Proposition 2.5. From {lim}_{x\to \mathrm{\infty}}G(x,a)=\mathrm{\infty}, there exists uniquely a number {x}_{3}>{x}_{2} satisfying G({x}_{3},a)=0. If 1<x<{x}_{3}, then G(x,a)<0, so {F}_{x}(x,a)<0. If x>{x}_{3}, then G(x,a)>0, so {F}_{x}(x,a)>0. Hence F is strictly decreasing for 1<x<{x}_{3} and strictly increasing for x>{x}_{3}. So, we get
F(x,a)\le max\{F(1,a),\underset{x\to \mathrm{\infty}}{lim}F(x,a)\}=1.
This completes the proof of Theorem 1.1. □
References
Cîrtoaje V: On some inequalities with powerexponential functions. JIPAM. J. Inequal. Pure Appl. Math. 2009., 10(1): Article ID 21
Cîrtoaje V: Proofs of three open inequalities with powerexponential functions. J. Nonlinear Sci. Appl. 2011, 4(2):130–137.
Acknowledgements
We would like to thank the referees for their valuable comments according to which various parts of this paper have been cleared and improved.
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MM had the first drift of the paper and YN added some contents and reorganized the paper. All authors read and approved the final manuscript.
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Miyagi, M., Nishizawa, Y. Proof of an open inequality with double powerexponential functions. J Inequal Appl 2013, 468 (2013). https://doi.org/10.1186/1029242X2013468
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DOI: https://doi.org/10.1186/1029242X2013468