JMO 2002

We need to prove this inequality: $\sum \frac{(b+c-a)^2}{(b+c)^2+a^2}\geq \frac{3}{5}$ $\Leftrightarrow \sum \frac{2ab+2ac}{a^2+b^2+c^2+2bc}\leq \frac{12}{5}$ Let $s=a^2+b^2+c^2$ $5s^2\sum ab+10s\sum a^2bc+20a^3b^2c\leq 6s^3+6s^2\sum ab+12s\sum a^2bc+48a^2b^2c^2\geq 0$ By applying Schur inequality: $\sum 4a^4bc-8a^3b^2c+4a^2b^2c^2\geq 0$ We prove that: $\sum 3a^6+2a^5b-2a^4b^2-a^4bc+2a^3b^3-4a^3b^2c^2\geq 0$ It's true because we have: $\frac{2\sum (2a^6+b^6)}{3}-a^4b^2\geq 0-(1)$ $\frac{\sum (4a^6+b^6+c^6)}{6}-a^4bc\geq 0-(2)$ $\frac{2\sum (2a^3b^3+c^3a^3) }{3}-a^2b^2c^2\geq 0-(3)$ $\frac{2\sum (2a^5b+a^5c+ab^5+ac^5)}{6}-a^3b^2c\geq 0-(4)$ From (1)+(2)+(3)+(4) we can solve the problem. The equality holds when a=b=c.

Without losing generality, we assume that $a+b+c=3$ , the expression changed to $\displaystyle\sum_{cyc}\frac{(3-2a)^2}{(3-a)^2+a^2}$ Now we prove this inequality $\frac{(3-2a)^2}{(3-a)^2+a^2}\geq\frac{18a-17}{5}$ $\Leftrightarrow \frac{18(a-1)(-a^2+5a+1)}{(3-a)^2+a^2}\geq 0 \ (True \ for \ every \ a\geq 0\ )$ Proving the similar inequality with $b$ and $c$ then combine all of them, we get $\displaystyle\sum_{cyc}\frac{(3-2a)^2}{(3-a)^2+a^2}\geq\frac{18(a+b+c)-51}{5}=\frac{3}{5}$ The equality holds when $a=b=c$