MetaMath '22 - demo

Primjenom $A-G$ nejednakosti na nazivnik prvog razlomka dobivamo $$\frac{2}{a+b} \leq \frac{2}{2\sqrt{ab}} = \frac{1}{\sqrt{ab}} = \frac{\sqrt{c}}{\sqrt{abc}}.$$ Analognom primjenom i na preostala $2$ nazivnika te sumiranjem dobivamo $$\sqrt{abc}\bigg( \frac{2}{a+b} + \frac{2}{b+c} + \frac{2}{c+a} \bigg) \leq \sqrt{abc}\bigg( \frac{\sqrt{c}}{\sqrt{abc}} + \frac{\sqrt{a}}{\sqrt{abc}} + \frac{\sqrt{b}}{\sqrt{abc}} \bigg) = \sqrt{a} + \sqrt{b} + \sqrt{c}.$$