TITLE: Prove that $3x-x^3<\frac2{\sin2x}$
QUESTION [6 upvotes]: Prove that $$3x-x^3<\frac2{\sin2x},\forall x\in\left(0,\frac\pi2\right)$$

I have tried by proving that $$3x-x^3<\frac9{5\pi}x+\frac32<\frac2{\sin2x},\forall x\in\left(0,\frac\pi2\right)$$ with Jensen's inequality, but I hoped this problem would have a simpler solution.

REPLY [4 votes]: HINT:
$$\sin 2x\le1 \implies \frac{2}{\sin 2x}\ge 2$$
and
$$3x-x^3\le 2$$

REPLY [2 votes]: The range of the function $f(x)=3x-x^3$ for $x\in(0,\pi/2)$ is $f(x)\in(0,2)$.
For the function $g(x)=\dfrac2{\sin2x}$ for $x\in(0,\pi/2)$ the range is clearly $g(x)\in(2,\infty)$.
The inequality is quite clear now.