I: 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 II: 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 $x=y=z=1$.

### 2.2 Combinatorics

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