Showing $\frac{y-x}{\cos{x}}(2+\sin{x}\frac{\cos{x}+\cos{y}+\cos{x}\cos{y}}{\sin{(y-x)}})\leq \pi $ I was playing around a bit with trigonometric inequalities and found this: 
$$\frac{y-x}{\cos{x}}(2+\sin{x}\frac{\cos{x}+\cos{y}+\cos{x}\cos{y}}{\sin{(y-x)}})\leq \pi
$$
for $0\leq x<y \leq\pi/2$.
This can be checked on Matlab for example. I know it's not the prettiest of inequalities but I found it interesting that the value $\pi$ is achieved for the pair $(x,y)= (0,\pi/2)$ and thought there could be some elegant way to show it.
I have tried various approaches with no luck. I think one promising approach is to show that the function is increasing with respect to $y$ when we fix $x$ but I found the calculations to be quite messy. I would appreciate any ideas on proving this.
 A: Fact 1: It holds that $z \le \frac{\pi \sin z}{2 + \cos z}$ for $z\in [0, \frac{\pi}{2}]$.
From Fact 1, it suffices to prove that
$$\frac{\pi \sin (y-x)}{2 + \cos (y-x)}\frac{1}{\cos x}\Big(2 + \sin x \frac{\cos x + \cos y + \cos x \cos y}{\sin (y-x)}\Big) \le \pi$$
which is equivalent to
$$[\cos^2 x + (1-\cos x)\sin x]\cos y + (2-\sin x)(1-\sin y)\cos x \ge 0.$$
Clearly, it is true. We are done.
$\phantom{2}$
Proof of Fact 1: It suffices to prove that $\pi \sin z - z \cos z - 2 z \ge 0$ for $z \in [0, \frac{\pi}{2}]$.
Let $f(z) = \pi \sin z - z \cos z - 2 z$. 
We have
$f'(z) = \pi \cos z - \cos z + z\sin z - 2$, $f''(z) = -\pi \sin z + 2 \sin z + z\cos z $
and $f'''(z) = -\pi \cos z + 3 \cos z - z \sin z$.
Clearly, $f'''(z) < 0$ for $z \in [0, \frac{\pi}{2}]$.
Also, $f''(0) = 0$. Thus, $f''(z) \le 0$ for $z \in [0, \frac{\pi}{2}]$.
Thus, $f(z)$ is concave on $[0, \frac{\pi}{2}]$.
Since $f(0) = f(\frac{\pi}{2}) = 0$, we have, for $z \in [0, \frac{\pi}{2}]$,
$$f(z) = f\Big((1-\tfrac{z}{\pi/2})\cdot 0 + \tfrac{z}{\pi/2} \cdot \tfrac{\pi}{2}\Big)
\ge (1-\tfrac{z}{\pi/2})f(0) + \tfrac{z}{\pi/2} f(\tfrac{\pi}{2}) = 0.$$ 
We are done.
