Question 74
Let $x$, $y$, $z$ be positive real numbers such that
$$
\sqrt{x}, \sqrt{y}, \sqrt{z} \text{ are sides of a triangle and } \frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 5.
$$
Prove that
$$
\frac{x(y^2 - 2z^2)}{z} + \frac{y(z^2 - 2x^2)}{x} + \frac{z(x^2 - 2y^2)}{y} \ge 0.
$$
$$
\sqrt{x}, \sqrt{y}, \sqrt{z} \text{ are sides of a triangle and } \frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 5.
$$
Prove that
$$
\frac{x(y^2 - 2z^2)}{z} + \frac{y(z^2 - 2x^2)}{x} + \frac{z(x^2 - 2y^2)}{y} \ge 0.
$$