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?
 A: $\dfrac{a^3}{b^2-bc+c^2}+\dfrac{b^3}{a^2-ac+c^2}+\dfrac{c^3}{a^2-ab+b^2}\geq a+b+c \iff \dfrac{a^3(b+c)}{b^3+c^3}+\dfrac{b^3(a+c)}{a^3+c^3}+\dfrac{c^3(b+a)}{b^3+a^3} \geq a+b+c \iff a^9b+a^9c+b^9a+b^9c+c^9a+c^9b \ge a^7b^3+a^7c^3+b^7a^3+b^7c^3+c^7a^3+c^7b^3$
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.
A: $$\Longleftrightarrow \dfrac{a^4}{ab^2-abc+ac^2}+\dfrac{b^4}{bc^2-abc+ba^2}+\dfrac{c^4}{ca^2-abc+cb^2}\ge a+b+c$$
by cauchy-Schwarz we have 
$$\left[ \dfrac{a^4}{ab^2-abc+ac^2}+\dfrac{b^4}{bc^2-abc+ba^2}+\dfrac{c^4}{ca^2-abc+cb^2}\right][bc(b+c)+ac(a+c)+ab(a+b)-3abc]\ge (a^2+b^2+c^2)^2$$
$$\Longleftrightarrow (a^2+b^2+c^2)^2\ge (a+b+c)[(bc(b+c)+ac(a+c)+ab(a+b)-3abc]$$
let $a=\min{\{a,b,c\}}$
$$\Longleftrightarrow a^2(a-b)(a-c)+(b^2+c^2+bc-ab-ac)(b-c)^2\ge 0$$
By Done
A: 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!
