Berkeley Math Circle Monthly Contest 5

Solution:

Note that $|x-y|$ is the greater of $x-y$ and $-(x-y)$. So the above assertion is equivalent to
$$\begin{aligned} & x+y > x-y \quad \text{ and } \quad x+y > -x+y \\ & 2y > 0 \quad \text{ and } \quad 2x > 0 \\ & y > 0 \quad \text{ and } \quad x > 0. \end{aligned}$$

Solution 1. The quantity
$$x + |x|$$
is $2x$, a positive number, if $x > 0$ and is $0$ otherwise. Therefore the inequality
$$(x + |x|)(y + |y|)(z + |z|) > 0$$
holds if and only if $x$, $y$, and $z$ are all positive.

Solution 2. By part (a), the condition that $x$, $y$, and $z$ are all positive is equivalent to
$$x + y - |x - y| > 0 \quad \text{ and } \quad z > 0.$$
By part (a) again, this is equivalent to
$$x + y - |x - y| + z > |x + y - |x - y| - z|.$$
This inequality involves no multiplication.