Question

ajtorr rapid la această

Answer 1

Folosim inegalitatea mediilor:
$m_a\geq m_g$

[tex]\frac{x^2+y^2}{2}\geq \sqrt{x^2y^2}\rightarrow \frac{x^2+y^2}{2}\geq xy\\ \frac{x^2+z^2}{2}\geq \sqrt{x^2z^2}\rightarrow \frac{x^2+z^2}{2}\geq xz\\ \frac{y^2+z^2}{2}\geq \sqrt{y^2z^2}\rightarrow \frac{y^2+z^2}{2}\geq yz\\ \rule{200}{0.5}+\\ \frac{2x^2+2y^2+2z^2}{2}\geq xy+xz+yz\\ \boxed{x^2+y^2+z^2\geq xy+xz+yz} [/tex]