I have proceeded(somewhat). Observe $\sqrt{xy+z}=\sqrt{(1-x)(1-y)}$ Now we can proceed using regular techniques or maybe by tigonometric substitutions.

$\sqrt{xy+z} = \sqrt{(1-x)(1-y)} = \sqrt{(z+y)(z+x)}$ So we want to prove that $\sum \sqrt{(z+y)(z+x)} \geq \sum x + \sum \sqrt{xy}$, but by Cauchy-Schwarz inequality we have $\sum \sqrt{(z+y)(z+x)} \geq \sum (z + \sqrt{xy}) = \sum x + \sum \sqrt{xy}$, QED.

OH!! That was trivial. I should have seen this there itself.

See also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=82809

zygfryd wrote: but by Cauchy-Schwarz inequality we have $ \sqrt{(z+y)(z+x)} \geq (z + \sqrt{xy}) $. Isn't this Huygen's inequality ?

utkarshgupta wrote: zygfryd wrote: but by Cauchy-Schwarz inequality we have $ \sqrt{(z+y)(z+x)} \geq (z + \sqrt{xy}) $. Isn't this Huygen's inequality ? Can you tell me what is Huygen's inequality? I can't find it anywhere

See here