, 2009

Solution:
Applying Cauchy-Schwarz's inequality:
$$\left(x^{2}+y+1\right)\left(z^{2}+y+1\right)=\left(x^{2}+y+1\right)\left(1+y+z^{2}\right) \geq (x+y+z)^{2}$$
Using the same reasoning we deduce:
$$\left(x^{2}+z+1\right)\left(y^{2}+z+1\right) \geq (x+y+z)^{2}$$
and
$$\left(y^{2}+x+1\right)\left(z^{2}+x+1\right) \geq (x+y+z)^{2}$$
Multiplying these three inequalities we get the desired result.

Solution:
We have
$$\begin{gathered} \left(x^{2}+y+1\right)\left(z^{2}+y+1\right) \geq (x+y+z)^{2} \Leftrightarrow \\ x^{2} z^{2}+x^{2} y+x^{2}+y z^{2}+y^{2}+y+z^{2}+y+1 \geq x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x \Leftrightarrow \\ \left(x^{2} z^{2}-2 z x+1\right)+\left(x^{2} y-2 x y+y\right)+\left(y z^{2}-2 y z+y\right) \geq 0 \Leftrightarrow \\ (x z-1)^{2}+y(x-1)^{2}+y(z-1)^{2} \geq 0 \end{gathered}$$
which is correct.
Using the same reasoning we get:
$$\begin{aligned} & \left(x^{2}+z+1\right)\left(y^{2}+z+1\right) \geq (x+y+z)^{2} \\ & \left(y^{2}+x+1\right)\left(z^{2}+x+1\right) \geq (x+y+z)^{2} \end{aligned}$$
Multiplying these three inequalities we get the desired result. Equality is attained at $x=y=z=1$.