Problem 1036

Prove that for positive numbers $x, y, z$,
$$\frac{z}{x y}(x-y)^{2}+\frac{x}{y z}(y-z)^{2}+\frac{y}{z x}(z-x)^{2} \geqslant 0,$$

By transformation, we can get $\frac{y z}{x}+\frac{z x}{y}+\frac{x y}{z} \geqslant x+y+z$
Let $x=b+c, y=c+a, z=a+b$, substitute into the inequality
(3), we get
$$\begin{array}{l} \quad \frac{(c+a)(a+b)}{b+c}+\frac{(a+b)(b+c)}{c+a}+ \\ \frac{(b+c)(c+a)}{a+b} \geqslant 2(a+b+c), \\ \quad \text { i.e. } \frac{a^{2}+b c+a(b+c)}{b+c}+\frac{b^{2}+c a+b(c+a)}{c+a}+ \\ \frac{c^{2}+a b+c(a+b)}{a+b} \geqslant 2(a+b+c), \end{array}$$

Thus, $\frac{a^{2}+b c}{b+c}+\frac{b^{2}+c a}{c+a}+\frac{c^{2}+a b}{a+b} \geqslant a+b+c$.