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A4

Part of 2011 Balkan MO Shortlist

Problems(1)

Inequality with condition xyz=3(x+y+z)

Source: Balkan MO ShortList 2011 A4

4/6/2020
Let x,y,zāˆˆR+x,y,z \in \mathbb{R}^+ satisfying xyz=3(x+y+z)xyz=3(x+y+z). Prove, that \begin{align*} \sum \frac{1}{x^2(y+1)} \geq \frac{3}{4(x+y+z)} \end{align*}
inequalitiesalgebraHigh school olympiadBalkan