Hard problem : Prove that $\Gamma\Big(\frac{1}{2}\Gamma\Big(\frac{1}{2}\Gamma\Big(\frac{1}{2}\Big)\Big)\Big)<\frac{\pi^2}{6}$ It's a problem that I can't solve it's :


$$\Gamma\Big(\frac{1}{2}\Gamma\Big(\frac{1}{2}\Gamma\Big(\frac{1}{2}\Big)\Big)\Big)<\frac{\pi^2}{6}$$


You have the difference here.
What I know :
We have $$\frac{\pi^2}{6}=\Gamma\Big(\frac{-3}{2}\Big)\Gamma^3\Big(\frac{3}{2}\Big)$$
Unfortunately we have a negative term so we cannot use the fact that  the gamma function is a strictly logarithmically convex function .
I have tried furthermore power series we have (see wiki page on Gamma function) :
$$\Gamma(z)=\frac{1}{z}-\gamma+\frac{1}{2}\Big(\gamma^2+\frac{\pi^2}{6}\Big)z+O(z^2)$$
But it's not sufficient and with more terms it's more delicate .
Finally I have tried Gautschi's inequality without success .
Now I think it's impossible to do it by hand and if someone do that I will open a bounty for him . 
Any helps is greatly appreciated...
...Thanks a lot for all your contributions.  
Update :
Recalling that :
$$\psi'(1)=\frac{\pi^2}{6}$$
Where $\psi(x)$ is the Digamma function .
We study the following function :
$$f(x)=\psi'(x)-\Gamma\Big(x\frac{1}{2}\Gamma\Big(\frac{1}{2}\Gamma\Big(\frac{1}{2}\Big)\Big)\Big)$$
We see that one of the roots of $f(x)$ is very near to one (wich is our inequality).So we can apply the Newton's method . 
Remains to show that the function is decreasing around one and evaluate $\frac{1}{2}\Gamma\Big(\frac{1}{2}\Gamma\Big(\frac{1}{2}\Big)\Big)$.
 A: Firstly,
$$\frac12\Gamma\left(\frac12\right)=\dfrac{\sqrt\pi}2\approx 0.886.\tag1$$
Denote
$$y=\frac12\Gamma\left(\frac12\Gamma\left(\frac12\right)\right),\tag2$$
then
$$y = \frac12\Gamma\left(\dfrac{1-x}{1+x}\right) < \frac12 + \gamma x + \left(\frac{\pi^2}6+\gamma^2-\gamma\right)x^2 + 0.251 x^3 + 2.33 x^4\tag3$$
(see also Wolfram Alpha series and coefficients),
where $\gamma$ is the Euler-Mascheroni constant,
$$x=\dfrac{2-\sqrt\pi}{2+\sqrt\pi}\approx0.060317809,\quad
\gamma\approx0.577215665.\tag4$$
The given inequality can be written in the form of
$$\dfrac1{\sqrt\pi}\Gamma(y) < \dfrac{\pi\sqrt\pi}6\approx0.928054666,\tag5$$
wherein
$$\dfrac1{\sqrt\pi}\Gamma(y) < 1 + \psi\left(\frac12\right)\left(y - \dfrac12\right) + \frac14\left(\pi^2+\psi^2\left(\frac12\right)\right)\left(y - \dfrac12\right)^2,\tag6$$
(see also Wolfram alpha series and coefficients),
where $\psi(x)$ is digamma function,
$$\psi\left(\frac12\right) = -\gamma - 2\ln2 \approx-1.963510026.\tag7$$
From $(3),(4)$ should
$$y-\dfrac12 < \frac1{25}.\tag8$$
From $(6)-(8)$ should
$$\dfrac1{\sqrt\pi}\Gamma(y) < 1 + \dfrac1{25}\psi\left(\frac12\right) + \dfrac1{2500}\left(\pi^2+\psi^2\left(\frac12\right)\right)\approx0.926949589
< \dfrac{\pi\sqrt\pi}6,$$
i.e. inequality $(5)$ is correct.
Therefore,
$$\color{brown}{\mathbf{\Gamma\left(\frac12\Gamma\left(\frac12\Gamma\left(\frac12 \right)\right)\right) < \frac{\pi^2}6}}.$$
