17th Turkish Mathematical Olympiad

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

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