Nice inequality ,Prove that $\Gamma\Big(\frac{\sin(x)}{x}\Big)\leq \frac{\pi}{\pi-x}$ Playing with geogebra I get :


Let $0\leq x<\pi$ then we have :
$$\Gamma\Big(\frac{\sin(x)}{x}\Big)\leq \frac{\pi}{\pi-x}$$
Where we have the Gamma function .


I have tried to use the Wendel inequality to prove that the ratio of the LHS and the RHS is one when $x$ tends to $\pi$ without success . The derivative is here but I can't handle it . I have tried power series of the Gamma function but I think it's reveals nothing good . So now I think it's not a trivial problem and I can't solve it .
My question :
How to solve it properly ?
Thanks in advance for your comments\answers.

Over $[0,\pi]$
$$ \Gamma\left(1+\frac{\sin x}{x}\right) \leq 1 \leq \frac{\pi \sin(x)}{x(\pi -x)}.$$
