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.
 A: Over $[0,\pi]$
$$ \Gamma\left(1+\frac{\sin x}{x}\right) \leq 1 \leq \frac{\pi \sin(x)}{x(\pi -x)}.$$
A: I do not know how much this could help you.
Since you played with Geogebra, you could check that the function
$$f(x)=\Gamma\left(\frac{\sin(x)}{x}\right)-\frac{\pi}{\pi-x}$$ can be quite nicely approximated by its $[1,1]$ Padé approximant built around $x=\frac \pi 2$
$$f(x) \sim \frac {a_0+a_1\left(x-\frac{\pi }{2} \right)} {1+a_2\left(x-\frac{\pi }{2} \right)}$$  The coefficients are quite nasty (have a look here) but numerically, they are
$$a_0 =-0.589172 \qquad a_1=-0.36822\qquad a_2=-0.164027$$
The maximum absolute error is $0.0086$ at $x=0$.
On the other hand, we also have
$$f(x)=-\frac{x}{\pi }+\left(\frac{\gamma }{6}-\frac{1}{\pi ^2}\right)
   x^2+O\left(x^3\right)$$
$$f(x)=(-1-\gamma )+\left(-\frac{\gamma ^2}{2 \pi }-\frac{\pi }{4}\right) (x-\pi
   )+\frac{(x-\pi )^2 \left(6 \gamma ^2-2 \gamma ^3-\pi ^2-\gamma  \pi ^2+2 \psi
   ^{(2)}(1)\right)}{12 \pi ^2}+O\left((x-\pi )^3\right)$$
