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Some remarks on Cîrtoaje’s conjecture
Journal of Inequalities and Applications volume 2016, Article number: 269 (2016)
Abstract
In this paper, we give new conditions under which the Cîrtoaje’s conjecture is also valid. We also show that a certain generalization of the Cîrtoaje’s inequality fulfils an interesting property.
1 Introduction and preliminaries
The study of inequalities with power-exponential functions is one of the active areas of research in the mathematical analysis. The power-exponential functions have useful applications in mathematical analysis and in other theories like statistics, biology, optimization, ordinary differential equations, and probability [1]. We note that the formulas of inequalities with power-exponential functions look so simple, but their solutions are not as simple as it seems. A lot of interesting results for inequalities with power-exponential functions have been obtained. The history and the literature review of inequalities with power-exponential functions can be found, for example, in [1]. Some other interesting problems concerning inequalities of power-exponential functions can be found in [2]. In this paper, we are studying one inequality conjectured by Cîrtoaje [3]. Cîrtoaje [3] has posted the following conjecture on the inequalities with power-exponential functions.
Conjecture 1.1
If \(a,b \in(0; 1]\) and \(r \in[0; e]\), then
The conjecture was proved by Matejíčka [4]. Matejíčka [5] also proved (1.1) under other conditions. Now we prove that the conjecture (1.1) is also valid under the following conditions:
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\(\frac{2}{e}\leq\min\{a,b\}\leq1\) and \(1\leq\max\{a,b\}\leq e\) for \(r \in[0; e]\);
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\(1\leq\min\{a,b\}\leq\max\{a,b\}\leq e\) for \(r \in[0; e]\).
We also show that a certain generalization of Cîrtoaje’s inequality fulfils an interesting property with some applications.
2 Main results
Theorem 2.1
Let \(a,b\) be positive numbers. Then
for any \(r\in[0,e]\) if one of the following two conditions is satisfied:
Proof
According to the proof of the Theorem 2.1 in [5], it suffices to consider the case where \(r=e\).
We split the proof into two parts, labeled as (a) and (b) with valid (2.2) and (2.3), respectively.
(a) Let a and b satisfy (2.2). Denote
for \(x \in[1,e]\). We have
where
and
If we show that \(H(1)\geq0\) and \(H'(x)\geq0\) for \(x \in[1,e]\), then the proof will be done.
To prove that \(H(1)\geq0\), we consider the function \(s: [2/e,1] \rightarrow\mathbf{R}\) defined as
We have that \(s(1)=0\). Now, if we show that \(s'(b)\leq0\) for \(b\in [2/e,1]\), then we can conclude that \(H(1)\geq0\). From
we obtain that \(s'(b)\leq0\) is equivalent to
Using
it suffices to show that
But the latter follows from \(v(2/e)=3-e+\frac{1}{2}\ln (\frac {e}{2} )>0\) and from \(v'(b)=(eb-1)/(2b)>0\).
Now we show \(F'(x)\geq0\) for \(x\in[1,e]\). This implies \(H'(x)\geq0\) for \(x\in[1,e]\). Indeed, if we show that \(F(1)\geq0\) and \(F'(x)\geq 0\), then \(F(x)\geq0\), so that \(H'(x)\geq0\).
We have
Because of \((\frac{eb}{2}-e)<-\frac{e}{2}\), it suffices to show that
for \(\frac{2}{e}\leq b\leq1\).
Since \(y(1)=0\), it suffices to show that
This is equivalent to
which follows from \(g(2/e)=0.8488\) and from
Indeed, \(g'<0\) follows from \(b\leq1< e/2\).
Next, we have that \(F'(x)\geq0\) is equivalent to
This can be rewritten as
Evidently,
So, to prove \(F'(x)\geq0\), it suffices to show that
Using \(\ln b>(b-1)/b\), we obtain
So we need to show that
Using again \(\ln b>(b-1)/b\) and \(\ln x>(x-1)/x\), it suffices to show that
Because of \(\ln x< x-1\), it suffices to prove that
It will be done if we show that \(r^{\ast\prime\prime}(x)\leq0\), \(r^{\ast }(1)\geq0\), and \(r^{\ast}(e)\geq0\).
We have
Because of \(r^{\ast\prime\prime}(x)=0\) only for one real root \(x_{1}=(eb^{2}-3b+2)/(2be-2e)<0\), we obtain \(r^{\ast\prime\prime}(x)\leq0\) for \(2/e\leq b<1\) and \(1\leq x\leq e\).
Now we show that \(r^{\ast}(e)\geq0\). We have
First, we show that \(u(1)\geq0\) and then \(u'(b)<0\). We have
Since
we obtain that \(u'(b)<0\) is equivalent to
It is evident that \(k(b)\) is a concave function. We show that \(k'=0\) only for \(m>1\) and \(k(1)<0\). This implies that \(k(b)<0\) for \(2/e\leq b\leq1\). So \(u'(b)<0\). Indeed, if \(k'=0\), then
We also have
Now we show that \(r^{\ast}(1)\geq0\). It will be done if we prove
But this follows from \(t'(b)\geq0\) and \(t(2/e)\geq0\). We have
and
The inequality \(t'(b)\geq0\) is equivalent to
since \(o(b)=2eb^{2}-2b+2e-2eb\geq0\), which is evident (\(o''(b)>0\), \(o'(b)=0\) for \(b=(1+e)/(2e)<2/e, o(2/e)>0\)). Now \(n(b)\geq0\) follows from \(n''(b)\geq0\), \(n'(b)=0\) for \(b=(5+e)/(2e)>1\), and \(n(1)=e^{2}-3e+2\doteq1.2342\geq0\).
(b) We assume that a and b satisfy (2.3).
We show again that \(H'(x)\geq0\) but now for \(1\leq b\leq x\leq e\). Because of \(H(b)=0\), the proof will be done.
From (2.4) we have that if \((eb-1-ex\ln b)\leq0\), then \(F'(x)\geq 0\). So we need to show that \(F'(x)\geq0\) for \(s(x,b)=(eb-1-ex\ln b)>0\).
Let \(s(x,b)=(eb-1-ex\ln b)>0\) for \(1\leq b\leq x\leq e\). Then \(F'(x)\geq0\) if only if
If \((eb-1-ex\ln b)>0\), then \(x<\frac{eb-1}{e\ln b}\). Because of \(x>b\), we have \(be\ln b< eb-1\). Put \(t=eb\) and \(v(t)=t\ln t-2t+1\) for \(e\leq t\leq e^{2}\). Then we have \(v(e)=1-e<0\), \(v(e^{2})=1\), \(v'(t)=\ln t-1>0\). This implies that there is only one \(t^{\ast}\) such that \(e< t^{\ast}< e^{2}\) and \(v(t^{\ast})=0\). Because of \(v(6.3055)=8.8113e-005>0\), we get \(t^{\ast}<6.3055\). So \(b< b^{\ast }<2.3196\). This implies that it suffices to show \(f(x,b)\geq0\) for \(1< b<2.3196\).
The mean value theorem gives
This implies
Similarly,
So
From \(\ln b>\frac{2(b-1)}{b+1}\) we have \(f(x,b)\geq G(x,b)\), where
We show that \(G(b,b)\geq0\) and \(G'_{x}(x,b)\geq0\), and the proof will be done.
We have
If we show that
then \(L(b)\geq0\), so \(G(b,b)\geq0\).
Inequality (2.7) is equivalent to
From \(s'(b)=2(e-2)b+(2-e)\), \(s'(b)=0\) if \(b=0.5\), \(s(1)=3-e>0\) we have \(s(b)>0\), so \(G(b,b)\geq0\).
Now we show \(G'_{x}(x,b)\geq0\) for \(1< b< x<\min\{e,(eb-1)/(e\ln b)\}\) and \(1< b< b^{\ast}\).
Because of \(eb-1-\frac{2ex(b-1)}{1+b}>eb-1-ex\ln(b)>0\), we have
(We omitted a positive term of derivation \(G'_{x}(x,b)\).)
Since
it suffices to show that
To prove (2.8), it suffices to show that (we used \(x>b\))
This can be rewritten as
where
From this we obtain that the roots of u are \(b_{1}=1.2468\) and \(b_{2}=5.4366\). We have that \(u<0\) on \((1,b_{1})\) and \(u>0\) on \((b_{1},b^{\ast})\). If we show that \(T(b,b)\geq0\), \(T(e,b)\geq0\) for \(b\in(1,b_{1})\) (\(T(x)\) is a concave function), and \(T'_{x}(x,b)\geq0\), \(T(b,b)\geq0\) for \(b\in(b_{1},b^{\ast})\), then the proof will be complete. Because of \(T'_{x}(x,b)=2xu+v\), it suffices to prove that \(P(b)=2bu(b)+v(b)\geq0\) for \(b\in(b_{1},b^{\ast})\).
First, we show \(T(b,b)\geq0\), \(T(e,b)\geq0\) for \(b\in(1,b_{1})\). We have
The roots of \(T(b,b)=0\) are \(r_{1}=-0.1913\), \(r_{2}=0.8517\), \(r_{3}=3.7046\), \(r_{4}=0\). This implies that \(T(b,b)\geq0\) for \(b\in(1,e)\).
Next, we have
The roots of \(T(e,b)=0\) are \(r_{1}=0.9969\), \(r_{2}=3.3956\), and \(T(e,0)<0\) implies \(T(e,b)\geq0\) for \(b\in(1,e)\).
Now we show that \(P(b)\geq0\) for \(b\in(b_{1},b^{\ast})\). We have
Because of \(P(b)=0\) has only one real root \(r_{1}=4.4344\), \(P(0)=18.5198\), and \(T(e,0)<0\), we obtain that \(p(b)\geq0\) for \(b\in (b_{1},b^{\ast})\).
So the proof is complete. □
3 Some generalizations of Conjecture 1.1
Denote \(M^{\ast}=\{(a,b); (0< a, b\leq e) \vee(0< b, a\leq e) \vee(a\geq e^{2}, b\leq\sqrt{a}) \vee(b\geq e^{2}, a\leq \sqrt{b}) \vee(0< a=b)\}\) (see Figure 1) and \(M(n,r)=\{(x_{1},\ldots,x_{n}); x_{i}>0, r\geq0 \wedge x_{1},\ldots,x_{n} \mbox{ are solutions of the}\mbox{ }\mbox{inequality (3.3)}\}\).
We have:
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\(\{(a,b); 0< a,b\leq e\}\subset M(2,e)\) (see [5]).
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\(\forall s>e\), \(\exists a,b<1\) such that \((a,b)\not\subset M(2,s)\) (see [4]).
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\(M^{\ast}\subset M(2,e)\).
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\((5,10)\not\subset M(2,e)\) (see [5]).
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\((1/3,1/9,2/3)\not\subset M(3,5/2)\) (see [5]).
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If \(0< r< s\), then \(M(n,s)\subset M(n,r)\) (Note 3.4).
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\(\forall x_{1},\ldots,x_{n}>0\), \(\exists s>0\) such that \((x_{1},\ldots,x_{n})\in M(n,r)\) for \(0\leq r\leq s\) (Note 3.4).
Lemma 3.1
[6], the log-sum inequality
Let \(n\in\mathbf {N}\), \(x_{1},\ldots,x_{n}, y_{1},\ldots,y_{n}\) be positive numbers. Then
with equality only for \(\frac{x_{1}}{y_{1}}=\frac {x_{2}}{y_{2}}=\cdots=\frac{x_{n}}{y_{n}}\).
Lemma 3.2
Let
where \(r\geq0\), \(n\in\mathbf{N}\), \(n\geq2\), \(x_{1},\ldots,x_{n}>0 \wedge \exists i\neq j\) such that \(x_{i}\neq x_{j}\). Then \(F(0)=0\), \(F'(0)>0\), \(F''(r)<0\).
Note 3.3
We note that \(F(r)\geq0\) is equivalent to
Proof
It is evident that \(F(0)=0\). Next, we have
The inequality \(F'(0)>0\) is equivalent to
which can be rewritten as
To prove (3.4), we use the Jensens log-sum inequality (Lemma 3.1).
Put \(y_{1}=x_{n}\), \(y_{2}=x_{1}, \dots,y_{n}=x_{n-1}\) in (3.1). We obtain
We show that \(v=v(y,x_{1},x_{n})>0\), where \(y=\sum_{i=2}^{n-1}x_{i}\). We have
It is evident that \(v(0,x_{1},x_{n})=(x_{n}-x_{1})\ln (\frac {x_{n}}{x_{1}} )>0\) and
If we show \(v'_{y}(y,x_{1},x_{n})\leq0\), then \(\lim_{y\rightarrow+\infty}v(y,x_{1},x_{n})=x_{n}-x_{1}+x_{1}\ln\frac {x_{1}}{x_{n}}\geq0\) (if we put \(t=x_{1}/x_{n}\), then \(g=1-t+t\ln t\geq 0\)) implies \(v(y,x_{1},x_{n})\geq0\).
Put \(t=\frac{y+x_{n}}{y+x_{1}}\) in (3.6). Then \(v'_{y}(y,x_{1},x_{n})=\ln t+1-t\). This implies \(v'_{y}(y,x_{1},x_{n})<0\).
Now we prove \(F''(r)<0\). We have
where
The equality \(L(r)\geq0\) can be rewritten as
From \(B_{2}\geq0\) and
we have \(A_{n}+B_{n}\geq0\) for \(n\geq2\). So, \(F(r)\) is a concave function for \(r\geq0\). □
Note 3.4
We note that Lemma 3.2 implies: if \(F(s)\geq0\) for some \(s>0\) and for positive numbers \(x_{1},\ldots,x_{n} \in M(n,s)\), then \(F(r)\geq0\) for \(r\in[0,s]\) on \(M(n,s)\).
3.1 Other applications of Lemma 3.2
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For each \(A\in R^{n}_{+}=\{(x_{1},\ldots,x_{n}), x_{i}>0, i=1,\ldots,n\}\), \(n\in\mathbf{N}\), there is a finite limit \(L_{A}=\lim_{r\rightarrow+\infty}F'(r)=\frac{1}{n}\sum_{i=1}^{n}x_{i}\log(x_{i})-m_{x}\), where \(m_{x}=\max_{1\leq m\leq n}\{x_{m+1}\log(x_{m})\}\), \(x_{n+1}=x_{1}\).
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Denote by \(r_{A}\) the positive root of \(F(r)=0\) (if the root exists) for \(A\in R^{n}_{+}\)-\(S^{n}\) where \(S^{n}=\{(x_{1},\ldots,x_{n}), x_{i}=x_{j}, i,j=1,\ldots,n\}\). Then
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(a)
\(L_{A}\geq0\) ⇔ there is no \(r_{A}>0\) such that \(F(r_{A})=0\).
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(b)
\(L_{A}<0\) ⇔ there is \(r_{A}>0\) such that \(F(r_{A})=0\).
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(a)
Let \(\emptyset\neq M\subset R^{n}_{+}-S^{n}\). Put \(r_{M}=\inf_{A\in M}\{r_{A}\}\) and \(R_{M}=\sup_{A\in M}\{r_{A}\}\). Then there are seven cases:
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(a)
\(r_{M}=R_{M}=0\),
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(b)
\(0=r_{M}< R_{M}<\infty\),
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(c)
\(r_{M}=0, R_{M}=\infty\),
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(d)
\(0< r_{M}=R_{M}<\infty\),
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(e)
\(0< r_{M}< R_{M}<\infty\),
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(f)
\(0< r_{M}< R_{M}=\infty\),
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(g)
\(r_{M}=R_{M}=\infty\).
From this we have:
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Case (a) is not possible (Lemma 3.2).
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In case (b), inequality (3.3) is not valid for \(r>0\) on M, but the reverse inequality to (3.3) is valid for \(r>R_{M}\) on M.
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In case (c), inequality (3.3) and the reverse inequality to (3.3) are not valid for \(r>0\) on M.
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In case (d), inequality (3.3) is valid for \(0\leq r\leq r_{M}\) on M, and the reverse inequality to (3.3) is valid for \(r>r_{M}\) on M.
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In case (e), inequality (3.3) is valid for \(0\leq r\leq r_{M}\) on M, but the reverse inequality to (3.3) is valid for \(r>R_{M}>r_{M}\) on M.
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In case (f), inequality (3.3) is valid for \(0\leq r\leq r_{M}<\infty\) on M, but the reverse inequality to (3.3) is not valid for any \(r>0\) on M.
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In case (g), inequality (3.3) is valid for all \(r\geq0\) on M.
3.2 Example
Let \(n=2\). Denote \(a=x_{2}\), \(b=x_{1}\). Then (1.1) is equivalent to \(F(r)\geq0\).
We have three cases:
Let
From \(b< a\) we have \(a\log(b)< b\log(a)\), so \((\frac{a-b}{2} )\log(a)+\frac{b}{2}\log (\frac{b}{a} )<0\). Lemma 2.2 in [4] gives that \(r_{M}=e\). \(\lim_{a\rightarrow1, b\rightarrow 0 }F(r)=\log2\) implies that \(R_{M}=\infty\). So, we have that the reverse inequality to (3.3) cannot be valid for any \(r>0\) on M.
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Acknowledgements
The work was supported by VEGA grant No. 1/0385/14. The author thanks Professor Vavro, Dean of the faculty FPT TnUAD in Púchov, Slovakia, for his kind support and is deeply grateful to the unknown reviewers for their valuable remarks and suggestions.
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Matejíčka, L. Some remarks on Cîrtoaje’s conjecture. J Inequal Appl 2016, 269 (2016). https://doi.org/10.1186/s13660-016-1211-0
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DOI: https://doi.org/10.1186/s13660-016-1211-0