How prove this inequality $\sum\limits_{cyc}\sqrt{\frac{yz}{x^2+2016}}\le\frac{3}{2}$ Given  $x,y,z$ are positive real number satisfy $xy+yz+xz=2016$. Prove that $$\sqrt{\dfrac{yz}{x^2+2016}}+\sqrt{\dfrac{xy}{z^2+2016}}+\sqrt{\dfrac{xz}{y^2+2016}}\le\dfrac{3}{2}$$

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 !

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, $$f(\sqrt{2016/3},\sqrt{2016/3},\sqrt{2016/3}) = 3/2.$$
