Problem

Source: Iran 3rd round 2011-algebra exam-p4

Tags: inequalities, function, LaTeX, Cauchy Inequality, inequalities proposed



For positive real numbers $a,b$ and $c$ we have $a+b+c=3$. Prove $\frac{a}{1+(b+c)^2}+\frac{b}{1+(a+c)^2}+\frac{c}{1+(a+b)^2}\le \frac{3(a^2+b^2+c^2)}{a^2+b^2+c^2+12abc}$. proposed by Mohammad Ahmadi