<s>[INST] Inequality olympiad For all positive numbers $a,b,c$, prove that 
$$\frac{a^3}{b^2-bc+c^2}+\frac{b^3}{a^2-ac+c^2}+\frac{c^3}{a^2-ab+b^2}\geq 3 \frac{(ab+bc+ac)}{a+b+c}$$
Note that both side are homogeneous of degree 1, so I think it is safe to assume $a+b+c=1$ but this does not go very far. 
Any ideas/hint?
Thanks
 [/INST] By Cauchy-Schwarz inequality:
$$\sum_{\text{cyc}}\dfrac{a^3}{b^2-bc+c^2}\left(\sum_{\text{cyc}}a\left(b^2-bc+c^2\right)\right)\ge \left(a^2+b^2+c^2\right)^2$$
In fact, you can prove the following stronger inequality:
$$\frac{\left(a^2+b^2+c^2\right)^2}{\sum_{\text{cyc}}a\left(b^2-bc+c^2\right)}\ge a+b+c\ge3\dfrac{ab+bc+ac}{a+b+c}$$
This holds:
$$\iff \left(a^2+b^2+c^2\right)^2\ge (a+b+c)\sum_{\text{cyc}}a\left(b^2-bc+c^2\right)$$
$$\iff a^4+b^4+c^4+abc(a+b+c)\ge ab\left(a^2+b^2\right)+bc\left(b^2+c^2\right)+ac\left(a^2+c^2\right)$$
The last step is true by Schur's inequality, where $t=2$.
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