Prove the inequality for $3$ numbers Given $x,y,z$ are reals such that $0<x<y<z<\frac{\pi}{2}$. Prove that $$\frac{\pi}{2}+2\sin x\cos y+2\sin y\cos z>\sin 2x+\sin 2y+\sin 2z$$ Since the inequality is not symmetric and equality too doesn't hold, I don't have many ideas. I tried using some rearrangement inequality, but it doesn't help. Thanks.

Our inequality is equivalent to:
$$f(x,y,z)=\sin(x)(\cos(x)-\cos(y))+\sin(y)(\cos(y)-\cos(z))+\sin(z)\cos(z)<\frac{\pi}{4}\tag{1}$$
but due to the concavity of the sine function, if we set $g(t)=\sin(t)\cos(t)$ we have:
$$\begin{eqnarray*}2\cdot f(x,y,z)&=&g(x)+g(z)+g(y-x)+g(z-y)\\&\leq&g(x)+g(z)+2\cdot g\left(\frac{z-x}{2}\right)\tag{2}\end{eqnarray*}$$
so it is enough to prove that if $0\leq x\leq z\leq\frac{\pi}{2}$,
$$ h(x,z)=\sin(x)\cos(x)+\sin(z)\cos(z)+\sin(z-x)\leq\frac{\pi}{2}\tag{3}$$
holds. It is easy to check that $\frac{\partial h}{\partial x}$ and $\frac{\partial h}{\partial z}$ both vanish iff $\cos(x+z)\cos(x-z)=0$, i.e. $z=\frac{\pi}{2}-x$. So, in order to prove $(3)$ it is enough to check that if $0\leq x\leq\frac{\pi}{4}$,
$$ j(x) = \sin(2x)+\cos(2x)\leq\frac{\pi}{2}\tag{4} $$
holds. That is trivial by the Cauchy-Schwarz inequality: moreover, the RHS of $(4)$ can be replaced with the smaller constant $\sqrt{2}$. So we have:

If $0\leq x\leq y\leq z\leq\frac{\pi}{2}$,
  $$\color{red}{\sqrt{2}}+2\sin x\cos y+2\sin y\cos z \geq \sin(2x)+\sin(2y)+\sin(2z).\tag{5}$$

