Turkish Mathematical Olympiad

$$\begin{align*} \sum_{\text{cyc}} \frac{(b+c)(a^4 - b^2c^2)}{ab+2bc+ca} &\ge \frac{1}{2} \sum_{\text{cyc}} (a^3 + abc - b^2c - bc^2) \\ &\ge \frac{1}{2} (a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b)) \ge 0 \end{align*}$$

by Schur's Inequality if $\frac{(b+c)(a^4 - b^2c^2)}{ab + 2bc + ca} \ge \frac{a^3 + abc - b^2c - bc^2}{2}$ for all positive real numbers $a, b, c$.
This last inequality is equivalent to $(b+c)a^4 - 2bca^3 - bc(b+c)a^2 + abc(b^2+c^2) \ge 0$, and since
$$(b+c)a^3 - 2bca^2 - bc(b+c)a + bc(b^2+c^2) \\ \ge \frac{4bc}{b+c}a^3 - 2bca^2 - bc(b+c)a + bc\frac{(b+c)^2}{2} \\ = \frac{bc}{2(b+c)}(2a+b+c)(2a-b-c)^2 \ge 0,$$
we are done.