Inegalitate conditionata cu x+y+z=1

[Claudiu Mindrila]

Daca \( x,y,z\in\mathbb{R}^*_+ \) astfel incat \( x+y+z=1 \), atunci \( \sqrt{\frac{x+yz}{yz}}+\sqrt{\frac{y+xz}{xz}}+\sqrt{\frac{z+xy}{xy}} \geq 6
\)

[Marius Mainea]

\( \sqrt{\frac{x+yz}{yz}}=\sqrt{\frac{x(x+y+z)}{yz}}=\sqrt{\frac{(x+y)(x+z)}{yz}}\ge\sqrt{\frac{2\sqrt{xy}\cdot2\sqrt{xz}}{yz}}=2\sqrt{\frac{x}{\sqrt{yz}}} \)
apoi se aplica inegalitatea mediilor pentru 3 numere.