Is there an elementary way to prove that $$\int\limits_{-1}^1 \frac{\arccos x}{\exp(x)+1}>\frac{\pi}{2} ~~?$$ The approximate value of the integral is 1.75.
I have tried to restrict to a smaller interval and to take minimum values of $\arccos$ on that interval.
By even/odd symmetry we have that
$$I \equiv \int_{-1}^1 \frac{\arccos x}{e^x+1}dx = \int_{-1}^1 \frac{\pi - \arccos x}{e^{-x}+1}dx$$
$$\implies I = \frac{\pi}{2}\int_{-1}^1 \frac{e^x}{e^x+1}dx + \frac{1}{2}\int_{-1}^1 \arccos x\cdot\frac{e^{-x}-e^x}{2+e^{-x}+e^x}dx$$
$$ = \frac{\pi}{2}+\frac{1}{2}\int_{-1}^1 \left(\frac{\pi}{2}-\arcsin x\right)\frac{e^{-x}-e^x}{2+e^{-x}+e^x}dx $$
$$ = \frac{\pi}{2}+\int_{0}^1 \arcsin x\cdot\frac{e^x-e^{-x}}{2+e^{-x}+e^x}dx$$
The integrand on the right is strictly positive, thus we have $I > \frac{\pi}{2}$