# Some remarks on Cîrtoaje’s conjecture

## 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.

## 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 . 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 . Some other interesting problems concerning inequalities of power-exponential functions can be found in . In this paper, we are studying one inequality conjectured by Cîrtoaje . Cîrtoaje  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

$$2\sqrt{a^{ra}b^{rb}}\geq a^{rb}+b^{ra}.$$
(1.1)

The conjecture was proved by Matejíčka . Matejíčka  also proved (1.1) under other conditions. Now we prove that the conjecture (1.1) is also valid under the following conditions:

• $$\frac{2}{e}\leq\min\{a,b\}\leq1$$ and $$1\leq\max\{a,b\}\leq e$$ for $$r \in[0; e]$$;

• $$1\leq\min\{a,b\}\leq\max\{a,b\}\leq e$$ for $$r \in[0; e]$$.

We also show that a certain generalization of Cîrtoaje’s inequality fulfils an interesting property with some applications.

## Main results

### Theorem 2.1

Let $$a,b$$ be positive numbers. Then

$$2\sqrt{a^{ra}b^{rb}}\geq a^{rb}+b^{ra}$$
(2.1)

for any $$r\in[0,e]$$ if one of the following two conditions is satisfied:

\begin{aligned}& \frac{2}{e}\leq b\leq1\leq a\leq e; \end{aligned}
(2.2)
\begin{aligned}& 1\leq b\leq a\leq e. \end{aligned}
(2.3)

### Proof

According to the proof of the Theorem 2.1 in , 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

$$H(x)=2\sqrt{x^{ex}b^{eb}}-x^{eb}-b^{ex}$$

for $$x \in[1,e]$$. We have

$$H'(x)=e\bigl(x^{\frac{ex}{2}}b^{\frac{eb}{2}}(\ln x+1)-bx^{eb-1}-b^{ex}\ln b\bigr)=eb^{ex}F(x),$$

where

$$F(x)=e^{\frac{e}{2} (x\ln x+b\ln b-2x\ln b )}(1+\ln x)-be^{(eb-1)\ln x-ex\ln b}-\ln b$$

and

\begin{aligned} F'(x) =& e^{\frac{e}{2} (x\ln x+b\ln b-2x\ln b )} \biggl(\frac {e}{2} (1+\ln x ) (1+ \ln x-2\ln b )+\frac{1}{x} \biggr) \\ &{} -be^{(eb-1)\ln x-ex\ln b} \biggl(\frac{eb-1}{x}-e\ln b \biggr). \end{aligned}

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

$$s(b)=H(1)=2b^{\frac{eb}{2}}-1-b^{e}.$$

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

$$s'(b)=e^{\frac{eb}{2}\ln b+1}(\ln b+1)-eb^{e-1}$$

we obtain that $$s'(b)\leq0$$ is equivalent to

$$\frac{eb\ln b}{2}-(e-1)\ln b+\ln(1+\ln b)\leq0 \quad\mbox{for } \frac{2}{e}\leq b\leq1.$$

Using

$$\ln(1+\ln b)\leq\ln b-\frac{\ln^{2}b}{2},$$

it suffices to show that

$$v(b)=\frac{eb}{2}+2-e-\frac{\ln b}{2}\geq0.$$

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

$$F(1)=e^{\frac{eb}{2}\ln b-e\ln b}-be^{-e\ln b}-\ln b.$$

Because of $$(\frac{eb}{2}-e)<-\frac{e}{2}$$, it suffices to show that

$$y(b)=b^{-\frac{e}{2}}-b^{1-e}-\ln b\geq0$$

for $$\frac{2}{e}\leq b\leq1$$.

Since $$y(1)=0$$, it suffices to show that

$$y'=-\frac{e}{2}b^{-\frac{e}{2}-1}-(1-e)b^{-e}- \frac{1}{b}\leq0.$$

This is equivalent to

$$g=-\frac{e}{2}b^{-\frac{e}{2}}+(e-1)b^{1-e}\leq1,$$

which follows from $$g(2/e)=0.8488$$ and from

$$g'=\frac{e^{2}}{4}b^{-\frac{e}{2}-1}-(e-1)^{2}b^{-e} \leq0.$$

Indeed, $$g'<0$$ follows from $$b\leq1< e/2$$.

Next, we have that $$F'(x)\geq0$$ is equivalent to

$$e^{\frac{e}{2} (x\ln x+b\ln b-2x\ln b )}\geq \frac{be^{(eb-1)\ln x-ex\ln b}(eb-1-ex\ln b)}{ \frac{ex}{2}(\ln x+1)(\ln x+1-2\ln b)+1}.$$
(2.4)

This can be rewritten as

\begin{aligned} &{\frac{e}{2} (x\ln x+b\ln b )}-(eb-1)\ln x \\ & \quad\geq\ln \bigl(2b(eb-1-ex\ln b) \bigr)-\ln \bigl(ex(\ln x+1) (\ln x+1-2\ln b)+2 \bigr). \end{aligned}
(2.5)

Evidently,

$$ex(\ln x+1) (\ln x+1-2\ln b)+2\geq ex+2.$$

So, to prove $$F'(x)\geq0$$, it suffices to show that

\begin{aligned} &{\frac{e}{2} (x\ln x+b\ln b )}-(eb-1)\ln x \\ &\quad \geq\ln \bigl(2b(eb-1-ex\ln b) \bigr)-\ln(ex+2). \end{aligned}

Using $$\ln b>(b-1)/b$$, we obtain

$$2b(eb-1-ex\ln b)< 2eb^{2}-2b+2ex(1-b).$$

So we need to show that

\begin{aligned} &{\frac{e}{2} (x\ln x+b\ln b )}-(eb-1)\ln x \\ &\quad \geq\ln \bigl(2eb^{2}-2b+2ex(1-b) \bigr)-\ln(ex+2). \end{aligned}

Using again $$\ln b>(b-1)/b$$ and $$\ln x>(x-1)/x$$, it suffices to show that

\begin{aligned} r(x)={}&\frac{e}{2} (x+b-2 )-(eb-1)\ln x \\ &{} -\ln \bigl(2eb^{2}-2b+2ex(1-b) \bigr)+\ln(ex+2)\geq0. \end{aligned}

Because of $$\ln x< x-1$$, it suffices to prove that

\begin{aligned} r^{\ast}(x)={}&\frac{e}{2} (x+b-2 )-(eb-1) (x-1) \\ &{}-\ln \bigl(2eb^{2}-2b+2ex(1-b) \bigr)+\ln(ex+2)\geq0. \end{aligned}

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

$$r^{\ast\prime\prime}(x)= \frac{4e^{2}(1-b)^{2}}{(2eb^{2}-2b+2ex(1-b))^{2}}-\frac {e^{2}}{(ex+2)^{2}}.$$

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

\begin{aligned} r^{\ast}(e)&= u(b) \\ &= \frac{e^{2}}{2}-1+b \biggl(\frac{3}{2}e-e^{2} \biggr)-\ln \bigl(2e^{2}-2b\bigl(1+e^{2} \bigr)+2eb^{2} \bigr)+\ln\bigl(e^{2}+2\bigr)\geq0. \end{aligned}

First, we show that $$u(1)\geq0$$ and then $$u'(b)<0$$. We have

$$u(1)=\frac{e^{2}}{2}-1+\frac{3}{2}e-e^{2}-\ln (2e-2 )+\ln \bigl(e^{2}+2\bigr)\doteq0.388>0.$$

Since

$$u'(b)=\frac{3e}{2}-e^{2}-\frac{4eb-2-2e^{2}}{2e^{2}-2b(1+e^{2})+2eb^{2}},$$

we obtain that $$u'(b)<0$$ is equivalent to

$$k(b)=2+2e^{2}+3e^{3}-2e^{4}-2b \bigl(1+e^{2}\bigr) \biggl(\frac{3e}{2}-e^{2} \biggr)-4eb+2e \biggl(\frac{3e}{2}-e^{2} \biggr)b^{2} \leq0.$$

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

$$m=\frac{(1+e^{2})(2e^{2}-3e)-4e}{2e(2e^{2}-3e)}=\frac {2e^{3}-3e^{2}+2e-7}{4e^{2}-6e}\doteq1.2411.$$

We also have

$$k(1)=2+2e^{2}+3e^{3}-2e^{4}-2b \bigl(1+e^{2}\bigr) \biggl(3\frac{e}{2}-e^{2} \biggr)-4eb+2e\biggl(3\frac {e}{2}-e^{2}\biggr)b^{2} \doteq-5.4757< 0 .$$

Now we show that $$r^{\ast}(1)\geq0$$. It will be done if we prove

$$t(b)=\frac{e}{2}(b-1)-\ln\bigl(2eb^{2}-2b+2e-2eb\bigr)+ \ln(e+2)\geq0.$$

But this follows from $$t'(b)\geq0$$ and $$t(2/e)\geq0$$. We have

$$t \biggl(\frac{2}{e} \biggr)=1-\frac{e}{2}-\ln \biggl( \frac{4}{e}+2e-4 \biggr)+\ln(e+2)\doteq0.1248\geq0$$

and

$$t'(b)=\frac{e}{2}-\frac{4eb-2e-2}{2eb^{2}-2b+2e-2eb}.$$

The inequality $$t'(b)\geq0$$ is equivalent to

$$n(b)=e^{2}b^{2}-b\bigl(5e+e^{2} \bigr)+e^{2}+2e+2\geq0$$

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

\begin{aligned} f(x,b)={}&{\frac{e}{2} (x\ln x+b\ln b )}-(eb-1)\ln x-\ln \bigl(2b(eb-1-ex\ln b) \bigr) \\ &{} +\ln \bigl(ex(\ln x+1) (\ln x+1-2\ln b)+2 \bigr)\geq0. \end{aligned}
(2.6)

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

$$\ln x-\ln b+1-\ln b >\frac{1}{x}(x-b)+\frac{1}{e}(e-b),\qquad \ln x+1\geq\frac{2x-1}{x}.$$

This implies

\begin{aligned} & \ln \bigl(ex(\ln x+1) (\ln x+1-2\ln b)+2 \bigr) \\ &\quad \geq\ln \bigl(e(2x-1) (2ex-eb-xb)+2ex \bigr)-\ln x-1. \end{aligned}

Similarly,

\begin{aligned} x\ln x+b\ln b-2b\ln x&=(x-b)\ln x+b(\ln b-\ln x)\geq(x-b) (\ln x-\ln e) \\ & >-\frac{(x-b)(e-x)}{x}. \end{aligned}

So

$$\frac{e}{2} (x\ln x+b\ln b )-eb\ln x\geq-\frac{e}{2} \frac {(x-b)(e-x)}{x}.$$

From $$\ln b>\frac{2(b-1)}{b+1}$$ we have $$f(x,b)\geq G(x,b)$$, where

\begin{aligned} G(x,b)={}&{-}\frac{e}{2}\frac{(x-b)(e-x)}{x}-\ln \biggl(2b \biggl(eb-1- \frac {2ex(b-1)}{1+b} \biggr) \biggr) \\ &{}+\ln \bigl((2x-1) (2ex-eb-xb)+2x \bigr). \end{aligned}

We show that $$G(b,b)\geq0$$ and $$G'_{x}(x,b)\geq0$$, and the proof will be done.

We have

\begin{aligned}& G(b,b)=-\ln \biggl(2b \biggl(eb-1-\frac{2eb(b-1)}{1+b} \biggr) \biggr)+\ln \bigl((2b-1) \bigl(eb-b^{2}\bigr)+2b \bigr)= \\& L(b)=\ln \biggl(\frac{1+b}{2} \biggr)-\ln \bigl(-eb^{2}+3eb-b-1 \bigr)+\ln \bigl(-2b^{2}+2eb+b-e+2 \bigr). \end{aligned}

If we show that

$$\frac{-2b^{2}+2eb+b-e+2}{-eb^{2}+3eb-b-1}\geq1,$$
(2.7)

then $$L(b)\geq0$$, so $$G(b,b)\geq0$$.

Inequality (2.7) is equivalent to

$$s(b)=(e-2)b^{2}+(2-e)b+3-e\geq0.$$

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

\begin{aligned} G'_{x}(x,b)\geq{}&{-}\frac{e}{2} \biggl( \frac{be-x^{2}}{x^{2}} \biggr)+\frac {2e(b-1)}{b ((eb-1)(1+b)-2ex(b-1) )} \\ &{} +\frac{8ex-4bx-2be-2e+b+2}{(2x-1)(2ex-eb-xb)+2x}. \end{aligned}

(We omitted a positive term of derivation $$G'_{x}(x,b)$$.)

Since

$$\frac{be-x^{2}}{x^{2}}< \frac{x(e-x)}{x^{2}}, \qquad\frac{2e(b-1)}{b ((eb-1)(1+b)-2ex(b-1) )}>0,$$

it suffices to show that

$$\frac{xe-e^{2}}{2x}+\frac{8ex-4bx-2be-2e+b+2}{(2x-1)(2ex-eb-xb)+2x}\geq 0.$$
(2.8)

To prove (2.8), it suffices to show that (we used $$x>b$$)

\begin{aligned} & x^{2}\bigl((4e-2b) \bigl(eb-e^{2}\bigr)+16e-8b\bigr) \\ &\quad{} +x\bigl(\bigl(eb-e^{2}\bigr) (2+b-2eb-2e)-4eb-4e+2b+4 \bigr)+eb\bigl(eb-e^{2}\bigr)\geq0. \end{aligned}

This can be rewritten as

$$T(x,b)=x^{2}u(b)+xv(b)+w(b)\geq0,$$

where

\begin{aligned} & u(b)=-2eb^{2}+b\bigl(6e^{2}-8\bigr)+e \bigl(16-4e^{2}\bigr), \\ & v(b)=b^{2}\bigl(e-2e^{2}\bigr)+b\bigl(2-2e-3e^{2}+2e^{3} \bigr)+2e^{3}-2e^{2}-4e+4, \\ & w(b)=e^{2}b^{2}-be^{3}. \end{aligned}

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

\begin{aligned} T(b,b)={}& b\bigl(-2eb^{3}+b^{2}\bigl(4e^{2}+e-8 \bigr)+b\bigl(2+14e-2e^{2}-2e^{3}\bigr) \\ &{}+4-4e-2e^{2}+e^{3}\bigr). \end{aligned}

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

$$T(e,b)=2e \bigl(b^{2}\bigl(-2e^{2}+e\bigr)+b \bigl(4e^{3}-2e^{2}-5e+1\bigr)-2e^{4}+e^{3}+7e^{2}-2e+2 \bigr).$$

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

$$P(b)=-4eb^{3}+b^{2}\bigl(10e^{2}+e-16\bigr)+b \bigl(-6e^{3}-3e^{2}+30e+2\bigr)+2e^{3}-2e^{2}-4e+4.$$

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. □

## 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:

• $$\{(a,b); 0< a,b\leq e\}\subset M(2,e)$$ (see ).

• $$\forall s>e$$, $$\exists a,b<1$$ such that $$(a,b)\not\subset M(2,s)$$ (see ).

• $$M^{\ast}\subset M(2,e)$$.

• $$(5,10)\not\subset M(2,e)$$ (see ).

• $$(1/3,1/9,2/3)\not\subset M(3,5/2)$$ (see ).

• If $$0< r< s$$, then $$M(n,s)\subset M(n,r)$$ (Note 3.4).

• $$\forall x_{1},\ldots,x_{n}>0$$, $$\exists s>0$$ such that $$(x_{1},\ldots,x_{n})\in M(n,r)$$ for $$0\leq r\leq s$$ (Note 3.4).

### Lemma 3.1

, the log-sum inequality

Let $$n\in\mathbf {N}$$, $$x_{1},\ldots,x_{n}, y_{1},\ldots,y_{n}$$ be positive numbers. Then

$$\sum_{i=1}^{n}x_{i}\ln \biggl( \frac{x_{i}}{y_{i}} \biggr)\geq \Biggl(\sum_{i=1}^{n}x_{i} \Biggr) \ln \biggl(\frac{\sum_{i=1}^{n}x_{i}}{\sum_{i=1}^{n}y_{i}} \biggr)$$
(3.1)

with equality only for $$\frac{x_{1}}{y_{1}}=\frac {x_{2}}{y_{2}}=\cdots=\frac{x_{n}}{y_{n}}$$.

### Lemma 3.2

Let

$$F(r)=\ln n+\frac{r}{n} \Biggl(\sum_{i=1}^{n}x_{i} \ln x_{i} \Biggr)-\ln \Biggl(e^{rx_{1}\ln x_{n}} +\sum _{i=1}^{n-1}e^{rx_{i+1}\ln x_{i}} \Biggr),$$
(3.2)

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

$$n\sqrt[n]{\prod_{i=1}^{n}x_{i}^{rx_{i}}} \geq x_{n}^{rx_{1}}+\sum_{i=1}^{n-1}x_{i}^{rx_{i+1}}.$$
(3.3)

### Proof

It is evident that $$F(0)=0$$. Next, we have

$$F'(r)=\frac{1}{n} \Biggl(\sum_{i=1}^{n}x_{i} \ln x_{i} \Biggr)-\frac {e^{rx_{1}\ln x_{n}}x_{1}\ln x_{n} +\sum_{i=1}^{n-1}e^{rx_{i+1}\ln x_{i}}x_{i+1}\ln x_{i}}{ e^{rx_{1}\ln x_{n}} +\sum_{i=1}^{n-1}e^{rx_{i+1}\ln x_{i}}}.$$

The inequality $$F'(0)>0$$ is equivalent to

$$\sum_{i=1}^{n}x_{i}\ln x_{i}-x_{1}\ln x_{n}-\sum _{i=1}^{n-1}x_{i+1}\ln x_{i}>0,$$

which can be rewritten as

$$\sum_{i=2}^{n}x_{i}\ln \biggl( \frac{x_{i}}{x_{i-1}} \biggr)+x_{1}\ln \biggl(\frac{x_{1}}{x_{n}} \biggr)>0.$$
(3.4)

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

$$\sum_{i=2}^{n}x_{i}\ln \biggl( \frac{x_{i}}{x_{i-1}} \biggr)+x_{1}\ln \biggl(\frac{x_{1}}{x_{n}} \biggr) \geq v= \Biggl(\sum_{i=2}^{n}x_{i} \Biggr) \ln \biggl(\frac{\sum_{i=2}^{n}x_{i}}{\sum_{i=1}^{n-1}x_{i}} \biggr)+x_{1}\ln \biggl( \frac{x_{1}}{x_{n}} \biggr).$$
(3.5)

We show that $$v=v(y,x_{1},x_{n})>0$$, where $$y=\sum_{i=2}^{n-1}x_{i}$$. We have

$$v(y,x_{1},x_{n})= (y+x_{n} )\ln \biggl( \frac {y+x_{n}}{y+x_{1}} \biggr)+x_{1}\ln \biggl(\frac{x_{1}}{x_{n}} \biggr).$$

It is evident that $$v(0,x_{1},x_{n})=(x_{n}-x_{1})\ln (\frac {x_{n}}{x_{1}} )>0$$ and

$$v'_{y}(y,x_{1},x_{n})=\ln \biggl( \frac{y+x_{n}}{y+x_{1}} \biggr)+\frac {x_{1}-x_{n}}{y+x_{1}}.$$
(3.6)

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

$$F''(r)=\frac{-L(r)}{ (\exp (rx_{1}\ln x_{n} )+\sum_{i=1}^{n-1}\exp (rx_{i+1}\ln x_{i} ) )^{2}}< 0,$$

where

\begin{aligned} L(r)={}& \Biggl(\exp (rx_{1}\ln x_{n} )x_{1}^{2} \ln^{2} x_{n}+\sum_{i=1}^{n-1} \exp (rx_{i+1}\ln x_{i} )x_{i+1}^{2} \ln^{2} x_{i} \Biggr) \\ &{}\times \Biggl(\exp (rx_{1}\ln x_{n} )+\sum _{i=1}^{n-1}\exp (rx_{i+1}\ln x_{i} ) \Biggr) \\ &{}- \Biggl(\exp (rx_{1}\ln x_{n} )x_{1}\ln x_{n}+\sum_{i=1}^{n-1}\exp (rx_{i+1}\ln x_{i} )x_{i+1}\ln x_{i} \Biggr)^{2}\geq0. \end{aligned}

The equality $$L(r)\geq0$$ can be rewritten as

\begin{aligned} L(r)={}&A_{n}+B_{n}=\sum_{i=1}^{n-1} \exp \bigl(r(x_{i+1}\ln x_{i}+x_{1}\ln x_{n}) \bigr) (x_{i+1}\ln x_{i}-x_{1} \ln x_{n})^{2} \\ &{}+\sum_{i=1}^{n-1}\sum _{j=1}^{n-1}\exp \bigl(r(x_{i+1}\ln x_{i}+x_{j+1}\ln x_{j} )\bigr) \bigl(x_{i+1}^{2} \ln^{2} x_{i}-x_{i+1}x_{j+1}(\ln x_{i})\ln x_{j} \bigr)\geq0. \end{aligned}

From $$B_{2}\geq0$$ and

$$B_{n+1}=B_{n}+\sum_{i=1}^{n-1} \exp \bigl(r(x_{i+1}\ln x_{i}+x_{n+1}\ln x_{n} )\bigr) (x_{i+1}\ln x_{i}-x_{n+1}\ln x_{n} )^{2}$$

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 $$M(n,s)$$.

### Other applications of Lemma 3.2

• 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 $$m_{x}=\max_{1\leq m\leq n}\{x_{m+1}\log(x_{m})\}$$, $$x_{n+1}=x_{1}$$.

• 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

1. (a)

$$L_{A}\geq0$$ there is no $$r_{A}>0$$ such that $$F(r_{A})=0$$.

2. (b)

$$L_{A}<0$$ there is $$r_{A}>0$$ such that $$F(r_{A})=0$$.

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:

1. (a)

$$r_{M}=R_{M}=0$$,

2. (b)

$$0=r_{M}< R_{M}<\infty$$,

3. (c)

$$r_{M}=0, R_{M}=\infty$$,

4. (d)

$$0< r_{M}=R_{M}<\infty$$,

5. (e)

$$0< r_{M}< R_{M}<\infty$$,

6. (f)

$$0< r_{M}< R_{M}=\infty$$,

7. (g)

$$r_{M}=R_{M}=\infty$$.

From this we have:

• Case (a) is not possible (Lemma 3.2).

• 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.

• In case (c), inequality (3.3) and the reverse inequality to (3.3) are not valid for $$r>0$$ on M.

• 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.

• 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.

• 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.

• In case (g), inequality (3.3) is valid for all $$r\geq0$$ on M.

### Example

Let $$n=2$$. Denote $$a=x_{2}$$, $$b=x_{1}$$. Then (1.1) is equivalent to $$F(r)\geq0$$.

We have three cases:

$$\left \{ \textstyle\begin{array}{@{}l@{\quad}l} b\log(a)>a\log(b); & \mbox{then } L_{A}=\lim_{r\rightarrow+\infty}F'(r)= (\frac{a-b}{2} )\log(a)+\frac {b}{2}\log (\frac{b}{a} );\\ a\log(b)>b\log(a); & \mbox{then } L_{A}=\lim_{r\rightarrow+\infty}F'(r)= (\frac{b-a}{2} )\log(b)+\frac {a}{2}\log (\frac{a}{b} );\\ b\log(a)=a\log(b); & \mbox{then } L_{A}=\lim_{r\rightarrow+\infty}F'(r)= (\frac{a-b}{2} )\log (\frac {a}{b} )\geq0. \end{array}\displaystyle \right .$$

Let

$$M=\bigl\{ (a,b); 0< b< a\leq1\bigr\} .$$

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  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.

## References

1. Coronel, A, Huancas, F: The proof of three power exponential inequalities. J. Inequal. Appl. 2014, 509 (2014). http://www.journalofinequalitiesandapplications.com/content/2014/1/509

2. Miyagi, M, Nishizawa, Y: A stronger inequality of Cîrtoaje’s one with power-exponential functions. J. Nonlinear Sci. Appl. 8, 224-230 (2015). http://www.tjnsa.com

3. Cîrtoaje, V: Proofs of three open inequalities with power exponential functions. J. Nonlinear Sci. Appl. 4(2), 130-137 (2011)

4. Matejíčka, L: Proof of one open inequality. J. Nonlinear Sci. Appl. 7, 51-62 (2014). http://www.tjnsa.com

5. Matejíčka, L: On the Cîrtoaje’s conjecture. J. Inequal. Appl. 2016, 152 (2016). doi:10.1186/s13660-016-1092-2

6. Cover, TM, Thomas, JA: Elements of Information Theory. Wiley, New York (1991)

## 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.

## Author information

Authors

### Corresponding author 