# Prove inequality $\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} ≤ 1.5$

Prove $\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} ≥ 1.5$ with $a + b + c = 3$ and $a,b,c > 0$ The correct question is $\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} ≥ 1.5$ (I have proved it)

Can prove $\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} \geq 1.5$ now: $\frac{a}{b^2 + 1} = a-\frac{ab^2}{b^2+1}\geq a-\frac{ab}{2}$ Do it the same for $\frac{b}{c^2 + 1}, \frac{c}{a^2 + 1}$, then sum... $\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} \geq a+b+c - \frac{ab+bc+ac}{2} = 3 - \frac{ab+bc+ac}{2}$. In different side: $(a+b+c)^2 \geq 3(ab+bc+ca) => ab+bc+ac \leq3$. So: $\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} \geq 3 - 1,5 = 1,5$

-
Show some work so we can best know where to help you. –  Jebruho Nov 5 '12 at 13:45
The supplemental answer was given by the OP, so I rolled back its removal. –  robjohn Nov 6 '12 at 16:50
The inequality does not hold! For $a = 2, b = c = 1/2$, we have $$\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} \geq \frac{a}{b^2 + 1} = \frac 8 5 > \frac 3 2$$