<s>[INST] Proving the inequality $\frac{a^3}{b^2-bc+c^2}+\frac{b^3}{a^2-ac+c^2}+\frac{c^3}{a^2-ab+b^2}\geq a+b+c$ I am trying to prove the following inequality
For all positive numbers $a$, $b$ and $c$ we have
$$\dfrac{a^3}{b^2-bc+c^2}+\dfrac{b^3}{a^2-ac+c^2}+\dfrac{c^3}{a^2-ab+b^2}\geq a+b+c$$
I can probably solve this by reducing it to Schur's inequality.
However, is there any other method?
 [/INST] $\dfrac{a^3}{b^2-bc+c^2}+\dfrac{b^3}{a^2-ac+c^2}+\dfrac{c^3}{a^2-ab+b^2}\geq a+b+c \iff \dfrac{a^3(b+c)}{b^3+c^3}+\dfrac{b^3(a+c)}{a^3+c^3}+\dfrac{c^3(b+a)}{b^3+a^3} \geq a+b+c \iff a^9b+a^9c+b^9a+b^9c+c^9a+c^9b \ge a^7b^3+a^7c^3+b^7a^3+b^7c^3+c^7a^3+c^7b^3$
lemma: $a^n+b^n\ge a^{n-1}b+b^{n-1}a \ge a^{n-2}b^2+b^{n-2}a^2 \ge ...$
$a^n+b^n\ge a^{n-1}b+b^{n-1}a \iff (a-b)(a^{n-1}-b^{n-1})\ge 0 \iff (a-b)^2(a^{n-2}+a^{n-3}b+a^{n-4}b^2+....ab^{n-3}+b^{n-2}) \ge 0$ 
which is true ,when $a=b$ the "=" holds.
so we have $a^9b+b^9a \ge a^8b^2+a^8b^2 \ge a^7b^3+b^7a^3$ 
QED.
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