Question Number 157071 by MathSh last updated on 19/Oct/21
$$\mathrm{if}\:\:\:\mathrm{0}<\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}}\:\:\:\mathrm{then}: \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\boldsymbol{\mathrm{x}}}\:+\:\frac{\mathrm{2}}{\pi\mathrm{x}}\:\leqslant\:\left(\mathrm{1}\:-\:\frac{\mathrm{2}}{\pi}\right)^{\mathrm{2}} +\:\frac{\mathrm{1}}{\mathrm{x}}\:+\:\frac{\mathrm{2}}{\pi} \\ $$