Gamma function inequality $\Gamma(x)\le\frac1x$

Question

Let $x>0$; I have to solve the following inequality $$ \Gamma(x)\le\frac1x $$ Now \begin{align*} \Gamma(x)\le\frac1x \Longleftrightarrow x\Gamma(x)-1\le0 \end{align*} But calling $F(x):=x\Gamma(x)$ the previous inequality holds iff $$ F(x)-F(1)\le0 $$

and here maybe I should exploit some convexity property; I know that the Gamma function (on $\Bbb R_{>0}$) is positive log-convex, thus it's convex. The identity is also convex (non strictly); and even though the product of two convex functions in not necessarely so, I think it's true when the two functions are non negative; thus $F$ would be convex, but from here I can't continue!

EDIT: Noticing that $x\Gamma(x)=\Gamma(1+x)$, we shoud prove that $\Gamma(1+x)\le1$, and looking at the graphof the Gamma, this is true iff $1\le1+x\le2$, i.e. $0\le x\le1$ (then we should consider $x\neq0$, since we're dealing with the Gamma function); but how can I prove this analitically?

Answer 1

You already have all the key ingredients:

1. $\log\Gamma$ is convex and $\Gamma$ is positive, hence $\Gamma$ is convex on $\mathbb{R}^+$;
2. Actually $\Gamma''(x)\geq c > 0$ for any $x\in\mathbb{R}^+$;
3. If $f(x)$ is a strictly convex and $C^2$ function on $\mathbb{R}^+$ the equation $f(x)=k$ cannot have more than two solutions; $\Gamma(x+1)=1$ has the solutions $x=0$ and $x=1$;
4. It follows that $\forall x\in(0,1)$ we have $\Gamma(x+1)<1$, as well as $\Gamma(x+1)>1$ for any $x>1$.

Answer 2

As $x>0$, your inequality is equivalent to $x \Gamma(x) \leq 1$, itself equivalent, using the functional equation of $\Gamma$ function, to:

$$\Gamma(x+1) \leq 1$$

But $\Gamma$ function is decreasing on $(0,x_0)$, then increasing on $(x_0,+\infty)$ with $\Gamma(1)=\Gamma(2)=1$ (see for example (https://math.stackexchange.com/q/3247), where $x_0$ is explicited as being $\approx 1.46$).

Thus (1) is true for $x+1$ outside interval $[1,2]$, Equivalently $x$ has to be outside $[0,1]$, i.e $x>1$.