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An inequality

Let x, y, z be positive real numbers such that xy + yz + zx = 2016. Prove that

    \[\sum\sqrt{\frac{yz}{x^2+2016}} \leq \frac{3}{2}\]

Solution

Applying AM – GM we have that

    \begin{align*} \sum \sqrt{\frac{xy}{z^2+2016}}&=\sum \sqrt{\frac{xy}{z^2+xy+xz+yz}}\\ &=\sum \sqrt{\frac{xy}{(x+z)(y+z)}} \\ &\leq \frac{1}{2}\sum \left(\frac{x}{x+z}+\frac{y}{y+z}\right)\\ &=\frac{1}{2}\sum \left(\frac{x}{x+z}+\frac{z}{z+x}\right)\\ &=\frac{3}{2} \end{align*}

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