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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 is going to be useful)

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up vote 5 down vote accepted

$ \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)} \geq \frac{(a^2+b^2+c^2)^2}{a^3+b^3+c^3+ab(a+b)+bc(b+c)+ca(c+a)} \tag{1} $


$\frac{{a_1}^2}{x_1}+\frac{{a_2}^2}{x_2}+\frac{{a_3}^2}{x_3} \geq \frac{(a_1+a_2+a_3)^2}{x_1+x_2+x_3} \tag{2}$

$ a^3+b^3+c^3+ab(a+b)+bc(b+c)+ca(c+a) $ $= a^3+b^3+c^3+a^{2}b+ab^2+b^{2}c+bc^2+c^{2}a+ca^2$ $= (a+b+c)(a^2+b^2+c^2) \tag{3}$

From $(1) $ and $(3)$

$ \frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2} $ $\geq \frac{a^2+b^2+c^2}{a+b+c}$ $\geq\frac{a^2}{a+b+c}+\frac{b^2}{a+b+c}+\frac{c^2}{a+b+c}$

Now applying $(2)$ again to the expression on the right

$ \frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}$ $\geq \frac{a^2}{a+b+c}+\frac{b^2}{a+b+c}+\frac{c^2}{a+b+c}$ $\geq \frac{(a+b+c)^2}{3(a+b+c)} = \frac{a+b+c}{3}$

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Let $S$ be the sum on the left hand side and $T$ be the sum:


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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