Generalization of the estimate $e^{x}\geq \dfrac{x^{n}}{n!}$

Question

It follows trivially from the definition of the exponential function that for every $x>0$ and $n\geq 1$, $$ e^{x}\geq\frac{x^n}{n!} = \frac{x^n}{\Gamma(n+1)}. $$ Is this inequality still true for real $n$? That is

Let $r>3$ be a real number. Is it true that $$ e^{x}\geq \frac{x^r}{\Gamma(r+1)} $$ for every $x\geq 1$?

Answer 1

The inequality holds for $r \ge 1$ and $x \ge 0$.

According to Pretty lower bound on the gamma function we have $$ \Gamma(r+1) \ge \left( \frac{r+1}{e}\right)^r \ge \left( \frac{r}{e}\right)^r $$ so that is suffices to show that $$ r^r e^{-r} \ge x^r e^{-x} \, , $$ which is straight-forward.

Answer 2

Note that $x^se^{-x}$ has a maximum w.r.t. $x$ at $x=s$, and hence it suffices to ask when we have

$$\Gamma(s+1)\stackrel?\ge s^se^{-s}=\left(\frac se\right)^s\ge x^se^{-x}$$

which is equivalent to asking when does this weak Stirling approximation give a lower bound to $\Gamma(s+1)$. A rigorous proof may be found by looking into the bounds from Laplace's method in the derivation of the Stirling approximation as shown on Wikipedia.