Answer to Question #152409 in Discrete Mathematics for Stan jacob

Question

How to prove that for all real numbers a,b and c that a² + 5b² + c² ≥ 4ab + 2bc?

Answer 1

$a^2+5b^2+c^2\geqslant4ab+2bc$

$a^2-4ab+5b^2+c^2-2bc\geqslant0$

$(a^2-4ab+4b^2)+(c^2-2bc+b^2)\geqslant0$

$(a-2b)^2+(c-b)^2\geqslant0$

the sum of two squares >= 0