Question Number 163588 by HongKing last updated on 08/Jan/22
$${In}\:\:\bigtriangleup{ABC}\:\:{prove}\:{that} \\ $$$$\frac{{a}}{{b}}\:+\:\frac{{b}}{{c}}\:+\:\frac{{c}}{{a}}\:+\:\frac{{R}^{\mathrm{2}} }{\mathrm{4}{r}^{\mathrm{2}} }\:\geqslant\:\mathrm{1}\:+\:\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\:\frac{{c}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:+\:\frac{{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} } \\ $$