Prove that $\frac {x \csc x + y \csc y}{2} < \sec \frac {x + y}{2}$ 
If $0 < x,y < \frac {\pi}{2}$, prove that:
$$
\frac {x \csc x + y \csc y}{2} < \sec \frac {x + y}{2}
$$

My attempt. First, I tried to change this inequality:
$$
\frac {\frac{x}{\sin x} + \frac{y}{\sin y}}{2} < \frac{1}{\cos \frac{x+y}{2}}
$$
Then,it's easy to know:
$$
LHS = \frac{1}{2} \left ( \frac {x}{2\sin \frac {x}{2} \cos \frac{x}{2}} + \frac{y}{2 \sin \frac{y}{2} \cos \frac{y}{2}} \right ) < \frac{1}{2} \left ( \frac{1}{\cos^{2} \frac{x}{2}} +\frac{1}{\cos^{2} \frac{y}{2}} \right )
$$
$$
RHS > \frac{1}{\cos \frac{x}{2} \cos \frac{y}{2}}
$$
How to solve it next? It seems this way is wrong.

Lemma 1. $f(x)=\frac{x}{\sin x}$ is a positive, increasing and convex function on $I=\left(0,\frac{\pi}{2}\right)$.
Let us assume that $\mu = \frac{x+y}{2}\in I$ is fixed and $x\leq y$. Let us set $\delta=\frac{y-x}{2}$. By Lemma 1,
$$ \sup_{\substack{0\leq \delta < \min\left(\mu,\frac{\pi}{2}-\mu\right)\\ }}\left(\frac{\mu-\delta}{\sin(\mu-\delta)}+\frac{\mu+\delta}{\sin(\mu+\delta)}\right)$$
is achieved at the right endpoint of the range for $\delta$. If $\mu\leq\frac{\pi}{4}$ such supremum equals $1+\frac{2\mu}{\sin(2\mu)}$.
If $\mu\geq\frac{\pi}{4}$ such supremum equals $\frac{\pi}{2}+\frac{2\mu-\pi/2}{\sin(2\mu-\pi/2)}$. The inequality
$$ \frac{\pi}{2}+\frac{2\mu-\pi/2}{\sin(2\mu-\pi/2)}\leq \frac{2}{\cos\mu} $$
over the interval $\left[\frac{\pi}{4},\frac{\pi}{2}\right)$ is very loose, hence the problem boils down to showing that
$$ 1+\frac{2\mu}{\sin(2\mu)} \leq \frac{2}{\cos\mu} $$
holds over the interval $\left(0,\frac{\pi}{4}\right)$. By multiplying both sides by $\sin\mu\cos\mu$ we get that the inequality is equivalent to
$$ \sin\mu\cos\mu + \mu \leq 2\sin\mu\tag{E} $$
which follows from the termwise integration of $1+\cos(2\mu)\leq 2\cos\mu$.
