Challenging problem :
Show that :
$$\frac{\gamma}{2-\gamma}\pi-\ln(3+\gamma)<0$$
I come up with this inequality with What is the limit $\lim_{x\to 0}\left(\frac{x!x!!!x!!!!!...}{x!!x!!!!x!!!!!!...}\right)^{-\frac{1}{x}}=^?$
I can show the inequalities :
$$\frac{\pi}{2}-1<\gamma<\frac{1}{\sqrt{3}}$$
Using the Leibniz formula (expansion of arctangent) and the first harmonic number using a computer . It doesn't help here
We can also use continued fraction for $\pi$ and for $\gamma$ see https://mathworld.wolfram.com/Euler-MascheroniConstantContinuedFraction.html
We can also continue to get :
$$\frac{\gamma}{2-\gamma}\pi-\ln\left(3+\gamma\right)+\sqrt{2}\ln\left(1+\left(1+\gamma-\frac{\pi}{2}\right)^{2}\right)-\frac{1}{2}\ln\left(1+\left(1+\gamma-\frac{\pi}{2}\right)^{3}\right)>0$$
How to show it by hand without a computer ?