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I want to prove that for all $x,r>0$, we have

$$\exp(x) \geq\frac{x^r}{\Gamma(r+1)}.$$

This is motivated by the fact that for any non-negative integer $m$, we have $$\exp(x)=\sum_{n=0}^\infty\frac{x^n}{n!}\geq\frac{x^m}{m!}=\frac{x^m}{\Gamma(m+1)}.$$

My observations:

  • By taking derivatives with respect to $x$, we see that it is enough to prove the inequality for all $0<r\le 1$.
  • I have not gotten anything useful by taking derivatives with respect to $r$, but maybe somebody is able to do more with this idea than me.
  • Numerically the inequality seems to hold with equality when $x,r\to0$.
Maximilian Janisch
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1 Answers1

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Note that your problem is equivalent to showing that

$$\Gamma(r+1)\ge\max_{x\ge0}\frac{x^r}{e^x}=\left(\frac re\right)^r$$

which is a weak version of the Stirling approximation. An elementary proof is given by

\begin{align}\Gamma(r+1)&=\int_0^\infty x^re^{-x}~\mathrm dx\\&>\int_0^rx^re^{-x}~\mathrm dx\\&=\left(\frac re\right)^r+\int_0^rrx^{r-1}e^{-x}~\mathrm dx\\&\ge\left(\frac re\right)^r\end{align}

Simply Beautiful Art
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