, 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: $2a^3 \ge (a^2+b^2)(2a-b) = 2a^3 + 2ab^2 - a^2b - b^3 \Leftrightarrow b(a-b)^2 \ge 0$.

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