For $a,b\in(0,\frac{\pi}{2}),a<b$, prove
$$\frac{1}{b-a}\int_a^b\frac{x}{\sin x}dx\leqslant\frac{a+b}{\sin a+\sin b}.$$
By mean value theorem, there is $c\in(a,b)$ s.t.
$$\frac{c}{\sin c}=\frac{1}{b-a}\int_a^b\frac{x}{\sin x}dx$$
I want to prove that
$$\frac{c}{\sin c}\leqslant\frac{a+b}{\sin a+\sin b}$$
for any $c\in (a,b)$. I am not sure if I am on the right track
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 $[a,b]$.
If $a=\pi/6$ and $b=\pi/3$ the right side is $(\sqrt{3}-1)\pi/2 \approx 1.1499.$ But $1 \in (\pi/6,\pi/3)$ with $1/\sin(1) \approx 1.18839.$ So another approach is needed. I think using the value of $c$ of the mean value theorem for integrals, one doesn't know enough about where $c$ lies to get accurate enough information for an inequality such as this.
