Prove that $(e+x)^{e-x}>(e-x)^{e+x}$ I get stuck with proving that $$(e+x)^{e-x}>(e-x)^{e+x}$$ for $x \in (0, e)$. All I know, is that it is doable with Jensen inequality, and I started with defining $$f(x)=(e+x)^{e-x}$$ and further $$g(x)=\ln \cdot f(x)$$ and... nothing more come to my mind, I kindly ask for any help & hints. Thanks
 A: Using power series,
$$
\begin{align}
(e-x)\log(e+x)
&=e-x+(e-x)\log\left(1+\frac xe\right)\\
&=e-x+(e-x)\left(\frac xe-\frac12\frac{x^2}{e^2}+\frac13\frac{x^3}{e^3}-\dots\right)\\
&=e-x+x-\frac32\frac{x^2}{e}+\frac56\frac{x^3}{e^2}-\dots\\
&=e-\frac32\frac{x^2}{e}+\frac56\frac{x^3}{e^2}-\frac7{12}\frac{x^4}{e^3}+\dots\tag{1}
\end{align}
$$
and
$$
\begin{align}
(e+x)\log(e-x)
&=e+x+(e+x)\log\left(1-\frac xe\right)\\
&=e+x-(e+x)\left(\frac xe+\frac12\frac{x^2}{e^2}+\frac13\frac{x^3}{e^3}+\dots\right)\\
&=e+x-x-\frac32\frac{x^2}{e}-\frac56\frac{x^3}{e^2}-\dots\\
&=e-\frac32\frac{x^2}{e}-\frac56\frac{x^3}{e^2}-\frac7{12}\frac{x^4}{e^3}-\dots\tag{2}
\end{align}
$$
Therefore,
$$
\begin{align}
&(e-x)\log(e+x)-(e+x)\log(e-x)\\
&=2\left(\frac56\frac{x^3}{e^2}+\frac9{20}\frac{x^5}{e^4}+\dots+\frac{4n+1}{2n(2n+1)}\frac{x^{2n+1}}{e^{2n}}+\dots\right)\tag{3}
\end{align}
$$
Thus, for $0\lt x\le e$ (so that the power series converge), this shows that
$$
(e+x)^{e-x}\gt(e-x)^{e+x}\tag{4}
$$
A: Here is an alternative to robjohn's nice answer.
This can be solved by an analysis of derivatives.
Hints: By dividing and then taking logs, the desired inequality is equivalent to showing $f(x) > 0$ on $(0,e)$ where
$$
f(x) := (e-x) \log (e+x) - (e+x) \log(e - x) \>.
$$
Steps:


*

*Verify that $f(0) = 0$.

*Verify that $f'(x) = -\log((e+x)(e-x)) + \frac{e+x}{e-x} + \frac{e-x}{e+x} > 0$ on the stated domain. 


To do this, use: 
Fact 1 The function $(e+x)(e-x)$ is a quadratic that takes the value $e^2$ at $x=0$ and the value $0$ at $x=e$, and,
Fact 2 For any $u > 0$, $v > 0$, we have $\frac uv + \frac v u = \frac{(u-v)^2 + 2uv}{uv} \geq 2$. 

A second simple variant of the above is to "pull out a multiplicative factor of $e$". Define $u = x/e \in (0,1)$ and note that the inequality in the question is equivalent to showing
$$
e^{e g(u)} > 1 \>,
$$
where $g(u) = (1-u) \log(1+u) - (1+u) \log (1-u) - 2 u$.
Again, show that $g(u) > 0$ by verifying that $g(0) = 0$ and $g'(u) > 0$ on $(0,1)$.
