TITLE: Prove that $(e+x)^{e-x}>(e-x)^{e+x}$
QUESTION [11 upvotes]: 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

REPLY [5 votes]: 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 $(0,1)$.