TITLE: How may I prove this integral inequality?
QUESTION [9 upvotes]: Prove the following inequality:
$$\frac{\sqrt{\pi}}{2}\le\int_{0}^{1}  \left({\log(\csc(x))}\right)^{1/3} dx$$
What should i start with? (it's not a homework but a hobby related activity)

REPLY [7 votes]: Note that $\sqrt{\pi}/2=\Gamma(3/2)$.
For $0<x\le 1$ we have $\csc(x)>1/x$ so
$$
\int_0^1 (\log(\csc(x)))^{1/3}~dx > \int_0^1(-\log(x))^{1/3}~dx
$$
Substitute $y=-\log(x)$ to get
$$
\int_0^1(-\log(x))^{1/3}~dx = \int_0^\infty y^{1/3}e^{-y}~dy=\Gamma(4/3)
$$
So your inequality follows from $\Gamma(4/3)>\Gamma(3/2)$. Unfortunately $\Gamma(4/3)$ does not have a simple expression, so I don't yet see a way to show this last step except by evaluating it numerically.