shortlistBMO 2011

The AM-GM inequality and the condition in the statement yield $x+y+z\ge 9$, so $$\frac{3}{4(x+y+z)}\sum (x+1)=\frac{3}{4}\left(1+\frac{3}{x+y+z}\right)\le 1.$$ Finally, apply the Cauchy-Schwarz inequality and take into account the condition in statement to get $$\left(\sum \frac{1}{x^2(y+1)}\right)\left(\sum (y+1)\right)\ge {\left(\sum \frac{1}{x}\right)}^2\ge 3\sum \frac{1}{xy}=1.$$ The conclusion follows. Clearly, equality holds if and only if $x=y=z=3$.