Problem 1628

12. Proof: (1) First, prove
$$2+x y z \geqslant \sum x$$

Since
$$1-y z \geqslant \frac{1}{2} \sum x^{2}-y z=\frac{1}{2} x^{2}+\frac{1}{2}(y-z)^{2} \geqslant 0$$

Similarly, we have
$$1-z x \geqslant 0,1-x y \geqslant 0$$

Therefore,
$$\begin{aligned} 2^{2}-\left(\sum x-x y z\right)^{2}= & 4-\sum x^{2}-2 \sum y z+2 x y z \sum x-(x y z)^{2} \geqslant \\ & 2-2 \sum y z+2 x y z \sum x-2(x y z)^{2}+(x y z)^{2}= \\ & 2 \prod(1-y z)+(x y z)^{2} \geqslant 0 \end{aligned}$$

That is,
$$\left(2-\sum x+x y z\right)\left(2+\sum x-x y z\right) \geqslant 0$$

From equation (2), to prove equation (1), it suffices to prove

Since
$$\begin{array}{c} 2+\sum x-x y z \geqslant 0 \\ 4 \geqslant 2 \sum x^{2} \geqslant 2\left(y^{2}+z^{2}\right) \geqslant(y+z)^{2} \\ 2 \geqslant|y+z| \geqslant-(y+z) \end{array}$$

Therefore,
That is,
$$2+y+z \geqslant 0$$

Since at least one of $x, y, z$ is non-negative, without loss of generality, assume $x \geqslant 0$, then
$$2+\sum x-x y z=(2+y+z)+x(1-y z) \geqslant 0$$
(Note that $1-y z \geqslant 0$ and equation (4)).
In summary, equation (1) is proved. Now use equation (1) to prove equation (1). Consider the following two cases.
i) When $x, y, z \leqslant 1, x \geqslant 0$, we have
$$x(1-y)(1-z) \geqslant 0$$

That is,

So
$$x+x y z \geqslant z x+x y$$

Additionally, we have
$$(1-x)(1-y z) \geqslant 0$$

That is
$$1-x+x y z \geqslant y z$$

By adding equations (5) and (6), we get equation (1).
ii) When one of $x, y, z$ is not less than 1, such as $x \geqslant 1$, at this time because
$$\begin{array}{c} x^{2}+y^{2}+z^{2} \leqslant 2 \quad x \geqslant 1 \\ y^{2}+z^{2} \leqslant 1 \\ |y| \leqslant 1,|z| \leqslant 1 \\ (1-x)(1-y)(1-z) \leqslant 0 \end{array}$$

So
So
That is
$$x y z+\sum x \geqslant 1+\sum y z$$

From the above proof, we have
$$2+x y z \geqslant \sum x$$

By adding equations (7) and (8), we get equation (1).
(2) Since
$$\begin{aligned} \sum y z-2 \sum x+3= & \frac{\left(\sum x\right)^{2}-\sum x^{2}}{2}-2 \sum x+3 \geqslant \\ & \frac{\left(\sum x\right)^{2}-2}{2}-2 \sum x+3= \\ & \frac{1}{2}\left(\sum x-2\right)^{2} \geqslant 0 \end{aligned}$$

Therefore, we get equation (2).