Question

Докажите неравенство 4bc<=4b2+c2

Answer 1

$4bc \leqslant 4 {b}^{2} + {c}^{2} \\ - 4 {b}^{2} +4bc - {c}^{2} \leqslant 0 \\ 4 {b}^{2} - 4bc + {c}^{2} \geqslant 0 \\ {(2b - c )}^{2} \geqslant 0$
Доказано!