Iranian Mathematical Olympiad

After the change of variables $(a, b, c) = (x + y, y + z, z + x)$ it suffices to prove that
$$a^2 + b^2 + c^2 + \frac{2ab}{1 + |a - b|} + \frac{2bc}{1 + |b - c|} + \frac{2ac}{1 + |c - a|} \geq 0$$
for all real numbers $a$, $b$, $c$. Let $(A, B, C) = (|b - c|, |c - a|, |a - b|)$, since $(ab) \cdot (bc) \cdot (ca) \geq 0$ we may assume that $ab \geq 0$. Since $A, B, C \geq 0$ and $A + B = |b - c| + |c - a| \geq |a - b| = C$, we have
$$\frac{ab}{1+C} \geq \frac{ab}{(1+A)(1+B)}$$
$$(a^2 + b^2 + \frac{2ab}{1+C} - (\frac{a}{1+B} + \frac{b}{1+A})^2) + (c + \frac{a}{1+B} + \frac{b}{1+A})^2 \geq 0$$

Second Solution. Since we can change $(a, b, c)$ by $(-a, -b, -c)$ we can assume that $a \geq b \geq 0 \geq c$. Then, if we put $C = |a - b|$, $A = |b - c|$, we can find $|a - c| = A + C$. Our inequality would indeed be
$$a^2 + b^2 + c^2 + \frac{2ab}{1+C} + \frac{2bc}{1+A} + \frac{2ac}{1+C+A} \geq 0$$
It is a quadratic expression with respect to $c$ and its discriminant is
$$(\frac{2b}{1+A} + \frac{2a}{1+C+A})^2 - 4(a^2 + b^2 + \frac{2ab}{1+C})$$
Then, $(\frac{2b}{1+A})^2 \leq 4b^2$, $(\frac{2a}{1+C+A})^2 \leq 4a^2$, and $2 \cdot \frac{2b}{1+A} \cdot \frac{2a}{1+C+A} \leq \frac{8ab}{1+C}$. Hence, the discriminant is non-positive and we are done. ■