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

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    $\begingroup$ take a root of $(e-x)(e+x)$ from both sides. Then check properties of $y=x^{1/x}$. $\endgroup$ – Maesumi Mar 3 '13 at 0:05
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    $\begingroup$ @Maesumi: I don't think that this works since $x^{1/x}$ is decreasing for $x>e$ and increasing for $x<e$. This shows that the maximum is at $x=e$, but comparison between one side of $e$ and the other is unclear. $\endgroup$ – robjohn Mar 3 '13 at 1:14
  • $\begingroup$ @robjohn I see my error now! $\endgroup$ – Maesumi Mar 3 '13 at 3:13

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 $$ (e+x)^{e-x}\gt(e-x)^{e+x}\tag{4} $$


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 $$ f(x) := (e-x) \log (e+x) - (e+x) \log(e - x) \>. $$


  1. Verify that $f(0) = 0$.
  2. 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 $(0,1)$.

  • $\begingroup$ A simplification is to note that for $0\lt x\lt e$, $$ \begin{align} f'(x) &=-\log(e^2-x^2)+2\frac{e^2+x^2}{e^2-x^2}\\ &\gt-2+2\\ &=0 \end{align} $$ (+1) $\endgroup$ – robjohn Mar 3 '13 at 18:49
  • $\begingroup$ @robjohn: That's a nice observation! I admit I stopped a little short since I wanted to leave something to the OP. :-) $\endgroup$ – cardinal Mar 3 '13 at 19:44

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