# Prove that 3 x &#x2212;<!-- − --> x 3 </msup> &lt; 2

Prove that
$3x-{x}^{3}<\frac{2}{\mathrm{sin}2x},\mathrm{\forall }x\in \left(0,\frac{\pi }{2}\right)$
I have tried by proving that
$3x-{x}^{3}<\frac{9}{5\pi }x+\frac{3}{2}<\frac{2}{\mathrm{sin}2x},\mathrm{\forall }x\in \left(0,\frac{\pi }{2}\right)$
with Jensen's inequality, but I hoped this problem would have a simpler solution.
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humusen6p
HINT:
$\mathrm{sin}2x\le 1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{2}{\mathrm{sin}2x}\ge 2$
and
$3x-{x}^{3}\le 2$
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flightwingsd2
The range of the function $f\left(x\right)=3x-{x}^{3}$ for $x\in \left(0,\pi /2\right)$ is $f\left(x\right)\in \left(0,2\right)$
For the function $g\left(x\right)=\frac{2}{\mathrm{sin}2x}$ for $x\in \left(0,\pi /2\right)$ the range is clearly $g\left(x\right)\in \left(2,\mathrm{\infty }\right)$
The inequality is quite clear now.