How prove this inequality $\sum\limits_{cyc}\sqrt{\frac{yz}{x^2+2016}}\le\frac{3}{2}$

Question

Given $x,y,z$ are positive real number satisfy $xy+yz+xz=2016$. Prove that $$\sqrt{\dfrac{yz}{x^2+2016}}+\sqrt{\dfrac{xy}{z^2+2016}}+\sqrt{\dfrac{xz}{y^2+2016}}\le\dfrac{3}{2}$$

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I tried

$\sqrt{\frac{yz}{x^2+2016}}=\sqrt{\frac{yz}{x^2+xy+yz+xz}}=\sqrt{\frac{yz}{\left(x+y\right)\left(x+z\right)}}$

And by C-S $\sqrt{\left(x+y\right)\left(x+z\right)}\ge x+\sqrt{yz}$

i can't continues. Help me !

Answer 1

By AM-GM $$\sum_{cyc}\sqrt{\frac{xy}{z^2+2016}}=\sum_{cyc}\sqrt{\frac{xy}{z^2+xy+xz+yz}}=\sum_{cyc}\sqrt{\frac{xy}{(x+z)(y+z)}}\leq$$ $$\leq\frac{1}{2}\sum_{cyc}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{x}{x+z}+\frac{z}{z+x}\right)=\frac{3}{2}$$

Answer 2

I believe there is something called the Purkiss Principle which would imply that in this case the maximum of $f$ is achieved when $x=y=z=\sqrt{2016/3}$. Thus, $$f(\sqrt{2016/3},\sqrt{2016/3},\sqrt{2016/3}) = 3/2.$$