Show that $\frac{\Gamma(1 + x)}{\Gamma(1 + x + n)} \leq \frac{1}{\Gamma(n+1)}$ for $x \in (0,1), n \in \mathbb{N}$

Question

Show that $\frac{\Gamma(1 + x)}{\Gamma(1 + x + n)} \leq \frac{1}{\Gamma(n+1)}$ for $x \in (0,1), n \in \mathbb{N}$

Here $\Gamma(\cdot)$ is the gamma function. This inequality is handy as $\Gamma(n+1) = n!$ which is a more elementary function.

Answer 1

Hint: Use $$\Gamma(y+1)=y\Gamma(y)$$ to establish $$\frac{\Gamma(1+x)}{\Gamma(1+x+n)}=\frac{1}{(x+1)(x+2)\cdots(x+n)}.$$ What happens when $x=0$?