Prove that:
$$\left(\int_1^{3}\frac{\ln^{3} x}{1+\sin x}\,dx\right)^{4}≤\left(\int_1^{3}\ln^{4} x\,dx\right)^{3}.$$
I need to use Hölder's or Chebyshev's inequality but I don't get it! $1+\sin x≤x+1$ so $\frac{1}{1+\sin x}≤1$.
Actually I don't know how to prove it!
The hint:
By Holder $$\left(\int\limits_1^3\ln^4xdx\right)^3\int\limits_1^3\frac{1}{(1+\sin{x})^4}dx\geq\left(\int\limits_1^3\frac{\ln^3x}{1+\sin{x}}\right)^4.$$ Thus, it's enough to prove that $$\int\limits_1^3\frac{1}{(1+\sin{x})^4}dx\leq1.$$ Now, since $f(x)=\frac{1}{1+4\sin{x}}$ is a convex function on $[1,3]$ and $\sin{x}>0$ for all $1\leq x\leq 3$, we obtain: $$\int\limits_1^3\frac{1}{(1+\sin{x})^4}dx\leq\int\limits_1^3\frac{1}{1+4\sin{x}}dx<\frac{\frac{1}{1+4\sin1}+\frac{1}{1+4\sin3}}{2}\cdot(3-1)=0.868...<1.$$