Ukrajina 2008

Let's consider the secondary inequality $\frac{a^3}{a^2+b^2}\ge a-\frac{b}{2}$ which is proved by means of rearrangement: $2a^3\ge (a^2+b^2)(2a-b)=2a^3+2a{b}^2-a^{2}b-b^3\iff b(a-b{)}^2\ge 0$.

$$\frac{a^2}{bc(a^2+b^2)}+\frac{b^2}{ca(b^2+c^2)}+\frac{c^2}{ab(c^2+a^2)}=\frac{1}{abc}\left(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\right)\ge \frac{1}{abc}\left(a-\frac{b}{2}+b-\frac{c}{2}+c-\frac{a}{2}\right)=\frac{a+b+c}{2abc}=\frac{(a+b+c)(ab+bc+ca)}{2abc}\ge \frac{3\sqrt{abc}\cdot \sqrt[3]{a^2{b}^2{c}^2}}{2abc}\ge \frac{9}{2}.$$