Question 28
Let $x$, $y$, $z$ be real numbers. Prove that:
$$
(x+y+z)^2 + \frac{(x+y)(y+z)}{1+|x-z|} + \frac{(y+z)(z+x)}{1+|y-x|} + \frac{(z+x)(x+y)}{1+|z-y|} \ge xy+yz+zx
$$
$$
(x+y+z)^2 + \frac{(x+y)(y+z)}{1+|x-z|} + \frac{(y+z)(z+x)}{1+|y-x|} + \frac{(z+x)(x+y)}{1+|z-y|} \ge xy+yz+zx
$$