How may I prove this integral inequality? 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)
 A: 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.
A: Proving such a thing with simple "analytic" inequalities seems hopeless, due to the combination of $\log$, $\csc$ and exponentiation by $1/3$- the integrand just seems too ugly for any simple "trick" to work. So we have to get our hands dirty. 
Calculating the derivative gives $\displaystyle \frac{-\cot x}{3 (\log (\csc(x)))^{2/3}}$ and since for $x\in [0,1]$ $\cot x \geq 0$ and $\csc (x) \geq 1$ we see the derivative is always $\leq 0$, i.e the function is decreasing. Thus taking right handed Riemann sums underestimates the integral. 
$$\int^1_0 \sqrt[3]{ \log(\csc (x))} dx > \frac{1}{20}\sum_{k=1} \sqrt[3]{ \log \left(\csc \left(\frac{k}{20}\right)\right)} = 0.915... $$
And since $\pi < 3.24 = 1.8^2$, we have $\displaystyle \frac{\sqrt{\pi}}{2} < 0.9$ so the desired inequality is reached. 
