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Algebra Difficulty 7.6 National olympiad, round 2 Prove it Saudi Arabia

For positive real numbers $a$ , $b$ and $c$ , prove that
$\frac{a^{3}}{a^{2}+a b+b^{2}}+\frac{b^{3}}{b^{2}+b c+c^{2}}+\frac{c^{3}}{c^{2}+c a+a^{2}} \geq \frac{a+b+c}{3}$

Solution

We have
$\begin{aligned} & \frac{a^{3}}{a^{2}+a b+b^{2}}+\frac{b^{3}}{b^{2}+b c+c^{2}}+\frac{c^{3}}{c^{2}+c a+a^{2}} \\ & \quad=\frac{a^{4}}{a^{3}+a^{2} b+a b^{2}}+\frac{b^{4}}{b^{3}+b^{2} c+b c^{2}}+\frac{c^{4}}{c^{3}+c^{2} a+c a^{2}} \\ & \quad \geq \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a^{3}+a b^{2}+a c^{2}+b a^{2}+b^{3}+b c^{2}+c a^{2}+c b^{2}+c^{3}} \end{aligned}$
by applying Cauchy-Schwarz inequality.

On the other hand
$\begin{aligned} \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a^{3}+a b^{2}+a c^{2}+b a^{2}+b^{3}+b c^{2}+c a^{2}+c b^{2}+c^{3}} & =\frac{a^{2}+b^{2}+c^{2}}{a+b+c} \\ & \geq \frac{a+b+c}{3} \end{aligned}$
by applying again Cauchy-Schwarz inequality.

The equality holds when $a=b=c$ .

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