Japan Mathematical Olympiad

From the Cauchy-Schwarz Inequality we get
$$\left(1 + \frac{y}{x} + \frac{z}{x}\right) \left(1 + xy + xz\right) \geq \left(1 + y + z\right)^2$$
Multiplying both sides of the inequality above by $\frac{x}{(x+y+z)(1+y+z)^2}$, we obtain
$$\frac{1 + xy + xz}{(1 + y + z)^2} \geq \frac{x}{x + y + z}.$$

Similarly, we obtain
$$\frac{1 + yz + yx}{(1 + z + x)^2} \ge \frac{y}{x + y + z}, \quad \frac{1 + zx + zy}{(1 + x + y)^2} \ge \frac{z}{x + y + z}$$
Adding both sides of these 3 inequalities side-by-side, we obtain the desired inequality:
$$\frac{1 + xy + xz}{(1 + y + z)^2} + \frac{1 + yz + yx}{(1 + z + x)^2} + \frac{1 + zx + zy}{(1 + x + y)^2} \ge 1$$