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

• The LHS terms can be rewritten as $\frac{a^3(b+c)}{b^3+c^3}$. Not sure if that would help. – Element118 Dec 27 '15 at 1:49
• @Element118, good caught, I made some calculation error – amathnerd Dec 27 '15 at 2:05

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 $t=2$.
• Why is $a+b+c\geq 3(\frac{ab+bc+ac}{a+b+c}$ – amathnerd Dec 27 '15 at 2:57
• @amathnerd $3(ab+bc+ca)\leq a^2+b^2+c^2+2ab+2bc+2ca=(a+b+c)^2$ – Element118 Dec 27 '15 at 3:16
• You mean "where $t=2$". I've edited it. – user236182 Dec 27 '15 at 7:22