<s>[INST] Prove for any positive real numbers $a,b,c$ $\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2} \geq \frac{a+b+c}{3}$ Since the problem sheets says I should use Cauchy-Schwarz inequality, I used
$\frac{{a_1}^2}{x_1}+\frac{{a_2}^2}{x_2}+\frac{{a_3}^2}{x_3}$ 
$\geq \frac{(a_1+a_2+a_3)^2}{x_1+x_2+x_3}$
I first multiplied each term by $a,b,c$ to get a perfect square on top like
$\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}$ 
$=\frac{a^4}{a(a^2+ab+b^2)}+\frac{b^4}{b(b^2+bc+c^2)}+\frac{c^4}{c(c^2+ca+a^2)}$
$\hspace{120pt}\geq \frac{(a^2+b^2+c^2)^2}{a^3+b^3+c^3+ab(a+b)+bc(b+c)+ca(c+a)} $
But I am still stuck for few hours. This is not a homework, but is a set of problems to prepare for AMC/USAMO.
(Note: I started off with a general question and got closed as off topic. I have another genuine Math Question now, and I am just trying to see if math.se is going to be useful) 
 [/INST] Let $S$ be the sum on the left hand side and $T$ be the sum:
$\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}$
Then $S-T = (a-b)+(b-c)+(c-a)=0$
So it suffices to show $S+T \geq \frac{2(a+b+c)}{3}$. We can achieve this by showing  $\frac{a^3+b^3}{a^2+ab+b^2} \geq \frac{a+b}{3}$
The inequality is equivalent to $a^2+b^2\geq 2ab$.
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