Prove $\frac{1}{b-a}\int_a^b\frac{x}{\sin x}dx\leqslant\frac{a+b}{\sin a+\sin b}$ $a,b\in(0,\frac{\pi}{2}),a<b$

Solution 1:

From the Hermite-Hadamard inequality, since $\frac{x}{\sin x}$ is convex on $[a,b]$, you have that $$\frac{1}{b-a}\int_a^b\frac{x}{\sin x}\,dx\leq\frac{1}{2}\left(\frac{a}{\sin a}+\frac{b}{\sin b}\right).$$ So it's enough to prove the following:

$$\frac{a\sin b+b\sin a}{2\sin a\sin a b}\leq\frac{a+b}{\sin a+\sin b}\Leftrightarrow(a+b)\sin a\sin b+b\sin^2a+a\sin^2 b\leq2(a+b)\sin a\sin b\Leftrightarrow b\sin^2a+a\sin^2b\leq(a+b)\sin a\sin b\Leftrightarrow$$$$b\sin a(\sin b-\sin a)+a\sin b(\sin a-\sin b)\geq 0\Leftrightarrow(\sin b-\sin a)(b\sin a-a\sin b)\geq 0.$$ But, $\sin b-\sin a>0$, so it is enough to show that $$b\sin a\geq a\sin b\Leftrightarrow\frac{\sin a}{a}\geq\frac{\sin b}{b}.$$ But, this holds, since $\frac{\sin x}{x}$ is decreasing in $[a,b]$.