Proof of a trigonometric inequality in $(0,\pi/2)$ I want to show
$$f(x)=-512\sin\frac{4x}7+1048\sin\frac{6x}7-800\sin\frac{8x}7+216\sin\frac{10x}7>0\quad x\in(0,\pi/2)$$
This trigonometric inequality has been verified by Mathematica using the Plot commend. However, I cannot give a rigorous proof of it. Any suggestion, idea, or comment is welcome, thanks!
 A: Let $\cos\frac{2x}{7}=t$.
Hence, $$f'(x)=\frac{16}{7}(t-1)^2(270t^3+140t^2-131t-34)$$
By the way, $t=\cos\frac{2x}{7}>\cos\frac{\pi}{7},$ which says that $f'(x)>0$
and $f(x)>f(0)=0$.
A: $$-512\sin\frac{4x}7+1048\sin\frac{6x}7-800\sin\frac{8x}7+216\sin\frac{10x}7>0,\ x\in(0,\pi/2)$$
$$\iff-512\sin2y+1048\sin3y-800\sin4y+216\sin5y>0,\ y\in(0,\pi/7)$$
$$\iff-64\sin2y+131\sin3y-100\sin4y+27\sin5y>0$$
Rewrite the LHS in terms of powers of $\sin y$ and $\cos y$:
$$\iff432\sin^5y+800\sin^3y\cos y-1064\sin^3y-528\sin y\cos y+528\sin y>0$$
$$\iff8\sin y\cdot(54\sin^4y+100\sin^2y\cos y-133\sin^2y-66\cos y+66)>0$$
$$\iff8\sin y\cdot(54(1-\cos^2y)^2+100(1-\cos^2y)\cos y-133(1-\cos^2y)-66\cos y+66)>0$$
$$\iff8\sin y\cdot(54\cos^4y-100\cos^3y+25\cos^2y+34\cos y-13)>0$$
$$\iff8\sin y\cdot(1-\cos y)^2(54\cos^2y+8\cos y-13)>0$$
$$\iff32\sin y\sin^4\frac y2(54\cos^2y+8\cos y-13)>0$$
$$\iff32\sin y\sin^4\frac y2(27\cos2y+8\cos y+14)>0$$
Since both $\sin y$ and $\sin\frac y2$ are strictly positive for $y\in(0,\pi/7)$:
$$\iff27\cos2y+8\cos y+14>0$$
Every term in the LHS is again strictly positive for $y\in(0,\pi/7)$. Therefore this last inequality is true, and following the implications back we see that the original inequality is true.
