Berkeley Math Circle: Monthly Contest 7

Solution:
The idea is to use the identity
$$\frac{x}{x+yz} = \frac{x}{x(x+y+z) + yz} = \frac{x}{(x+y)(x+z)} = \frac{x(y+z)}{(x+y)(y+z)(z+x)}$$
So the left-hand side is exactly equal to
$$\frac{2(xy+yz+zx)}{(x+y)(y+z)(z+x)} = \frac{2(xy+yz+zx)(x+y+z)}{(x+y)(y+z)(z+x)}.$$
We let $k = (xy+yz+zx)(x+y+z)$. On the other hand, we claim that $k \leq 1/3$, which is sufficient. Indeed, to prove $k \leq 1/3$ it suffices to prove that
$$xy+yz+zx \leq \frac{(x+y+z)^2}{3}$$
which is a direct consequence of the Cauchy-Schwarz inequality.

On the other hand, the left-hand side equals, exactly,
$$\frac{2k}{k-xyz} = 2 + \frac{2xyz}{k-xyz}$$
So simply noting $k \leq 1/3$ implies the desired conclusion, since this is a decreasing function of $k$.