gobathegreat wrote: Let $a$, $b$ and $c$ be positive real numbers such that $a^2+b^2+c^2=3$. Prove the following inequality: $\frac{a}{3c(a^2-ab+b^2)} + \frac{b}{3a(b^2-bc+c^2)} + \frac{c}{3b(c^2-ca+a^2)} \leq \frac{1}{abc}$ \[\sum \frac{a}{3c(a^2-ab+b^2)} \leq \sum \frac{a}{3abc }=\frac{a+b+c}{3abc} \leq \frac{\sqrt{3(a^2+b^2+c^2)}}{3abc}=\frac{1}{abc}\]

By AM-GM we have \begin{align*} a^2+b^2 &\geq 2ab \\ a^2-ab+b^2 &\geq ab \\ \frac{1}{a^2-ab+b^2} &\leq \frac{1}{ab} \end{align*}Multiplying $a/3c$ both side and cyclically summing gives \begin{align*} \sum_{\text{cyc}} \frac{a}{3c(a^2-ab+b^2)} &\leq \frac{1}{abc}\cdot \frac{a+b+c}{3} \\ &\stackrel{\text{QM-AM}}{\leq} \frac{1}{abc}\cdot \sqrt{\frac{a^2+b^2+c^2}{3}} \\ &\leq \frac{1}{abc} \end{align*}Equality holds when $a=b=c=1$.

By AM-GM we have that 1/(a^2+b^2) <= 1/2ab so we have that a/3c(a^2+b^2-ab) <= a/3abc we do the same for the other 2 inequalities, so now we have to show that cyc SUM of a <= cyc SUM of a^2 But this is obv from Cauchy that (a^2 + b^2 + c^2)(1 + 1 + 1) >= (a + b + c)^2 thus (cyc SUM of a^2)^2 >= (cyc SUM of a)^2 with equality for a=b=c=1