Question

problema 2 plz repede

Answer 1

$(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc=(6p)^2=36p^2$
$(a-b)^2 \geq 0$ ⇔ $a^2+b^2-2ab \geq 0$ ⇔ $2ab \leq a^2+b^2$
Analog se demonstreaza ca $2ac \leq a^2+c^2$
si
$2bc \leq b^2+c^2$

Revenim la prima egalitate scrisa si tinand cont de inegalitatile demonstrate pt produsele 2ab, 2ac si 2bc, obtinem:
$a^2+b^2+c^2+2ab+2bc+2ac \leq a^2+b^2+c^2+a^2+b^2+b^2+c^2+a^2+$$+c^2$

$36p^2 \leq 3(a^2+b^2+c^2)$
Impartind cu 3 ambii memri rezulta ceea ce trebuia demonstrat