Question

Demonstrati inegalitatea:
$\frac{a+b}{2} \geq \sqrt{ab}$

Answer 1

$\frac{a+b}{2} \geq \sqrt{ab}|()^{2} \Rightarrow \frac{(a+b)^{2} }{4} \geq ab \Rightarrow (a+b)^{2} \geq 4ab \Rightarrow a^{2} +2ab+ b^{2} \geq4ab \\ \Rightarrow a^{2} -2ab + b^{2} \geq 0 \Rightarrow (a-b)^{2} \geq 0$

Adevarat! Intotdeauna un numar real la patrat va avea ca rezultat un numar mai mare sau egal cu 0.