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?

We need to prove that
$$\sum_{cyc}\left(\frac{a^3}{b^2-bc+c^2}-a\right)\geq0$$ or
$$\sum_{cyc}\frac{a(a^2+bc-b^2-c^2)}{b^2-bc+c^2}\geq0$$ or
$$\sum_{cyc}\frac{a((a-b)(a+2b-c)-(c-a)(a+2c-b))}{b^2-bc+c^2}\geq0$$ or
$$\sum_{cyc}(a-b)\left(\frac{a(a+2b-c}{b^2-bc+c^2}-\frac{b(b+2a-c)}{a^2-ac+c^2}\right)\geq0$$ or
$$\sum_{cyc}(a-b)^2((a+b)^3-(a^2+3ab+b^2)c+2(a+b)c^2-c^3)(a^2-ab+b^2)\geq0.$$
Now, since by AM-GM $$(a+b)c^2-(a^2+3ab+b^2)c+(a+b)^3\geq2(a+b)^2c-(a^2+3ab+b^2)c>0,$$
it remains to prove that
$$\sum_{cyc}(a-b)^2c^2(a+b-c)(a^2-ab+b^2)\geq0.$$
Let $a\geq b\geq c$.
Hence, $$b^2\sum_{cyc}(a-b)^2c^2(a+b-c)(a^2-ab+b^2)\geq$$
$$\geq b^2(a-c)^2b^2(a+c-b)(a^2-ac+c^2)+b^2(b-c)^2a^2(b+c-a)(b^2-bc+c^2)\geq$$
$$\geq a^2(b-c)^2b^2(a+c-b)(a^2-ac+c^2)+b^2(b-c)^2a^2(b+c-a)(b^2-bc+c^2)\geq$$
$$\geq a^2(b-c)^2b^2(a-b)(a^2-ac+c^2)+b^2(b-c)^2a^2(b-a)(b^2-bc+c^2)=$$
$$=(a-b)^2(b-c)^2a^2b^2(a+b-c)\geq0.$$
Done!
