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.

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

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

(1.1)

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:

• $\frac{2}{e}\leq\min\{a,b\}\leq1$ 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.

2 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 [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

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

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. □

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 $M(n,r)=\{(x_{1},\ldots,x_{n}); x_{i}>0, r\geq0 \wedge x_{1},\ldots,x_{n} \mbox{ are solutions of the}\mbox{ }\mbox{inequality (3.3)}\}$.

Figure 1

Part of the set $\pmb{M^{\ast}}$ .

We have:

• $\{(a,b); 0< a,b\leq e\}\subset M(2,e)$ (see [5]).

• $\forall s>e$, $\exists a,b<1$ such that $(a,b)\not\subset M(2,s)$ (see [4]).

• $M^{\ast}\subset M(2,e)$.

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

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

• 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

[6], 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 $M(n,s)$.

3.1 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

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

  • (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:

• (a) $r_{M}=R_{M}=0$,

• (b) $0=r_{M}< R_{M}<\infty$,

• (c) $r_{M}=0, R_{M}=\infty$,

• (d) $0< r_{M}=R_{M}<\infty$,

• (e) $0< r_{M}< R_{M}<\infty$,

• (f) $0< r_{M}< R_{M}=\infty$,

• (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.

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:

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

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 [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.