<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] $$\Longleftrightarrow \dfrac{a^4}{ab^2-abc+ac^2}+\dfrac{b^4}{bc^2-abc+ba^2}+\dfrac{c^4}{ca^2-abc+cb^2}\ge a+b+c$$
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
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