Solving and proving inequalities? If $a,b$ and $c$ are positive real numbers, how do I prove that:
$$\frac{a^3}{b^2-bc+c^2}+
\frac{b^3}{c^2-ca+a^2} + \frac{c^3}{a^2-ab+b^2} \geq 3 \cdot
\frac{ab+bc+ca}{a+b+c}.$$
and when is equality?
Are there general techniques to solve these symmetric and or cyclic inequalities?
 A: Lemma 1: If $a,b$ and $c$ are positive real numbers, then $\frac{a^3}{b^2-bc+c^2}+\frac{b^3}{a^2-ac+c^2}+\frac{c^3}{a^2-ab+b^2}\geq a+b+c.$ 
Proof: Proving the inequality $\frac{a^3}{b^2-bc+c^2}+\frac{b^3}{a^2-ac+c^2}+\frac{c^3}{a^2-ab+b^2}\geq a+b+c$ $\blacksquare$
Lemma 2: If $a,b$ and $c$ are positive real numbers, then $a^2 + b^2 + c^2 ≥ ab + bc + ac.$ 
Proof: Is this correct method to prove that $a^2 + b^2 + c^2 ≥ ab + bc + ac$, when $a,b,c \geq 0$? $\blacksquare$
Claim: If $a,b$ and $c$ are positive real numbers, then $a+b+c ≥ 3 \cdot
\frac{ab+bc+ca}{a+b+c}.$ 
Proof: $a+b+c ≥ 3 \cdot
\frac{ab+bc+ca}{a+b+c}$
$\iff a^3+b^3+c^3 \geq 3abc$
$\iff (a+b+c) \cdot (a^2 + b^2 + c^2 - ab - bc - ac) \geq 0$
If $a+b+c=0$ then the above is trivially true.
Thus if $a+b+c>0$ we need to show that $a^2 + b^2 + c^2 ≥ ab + bc + ac$, which is true by Lemma 2. $\blacksquare$
Combining the Claim and Lemma 1 we prove your inequality that is $$ \frac{a^3}{b^2-bc+c^2}+
\frac{b^3}{c^2-ca+a^2} + \frac{c^3}{a^2-ab+b^2} \geq 3 \cdot
\frac{ab+bc+ca}{a+b+c}.$$
A: By C-S $$\sum_{cyc}\frac{a^3}{b^2-bc+c^2}=\sum_{cyc}\frac{a^4}{b^2a+c^2a-abc}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^2b+a^2c-abc)}.$$
Thus, it remains to prove that 
$$(a^2+b^2+c^2)^2(a+b+c)\geq3(ab+ac+bc)\sum\limits_{cyc}(a^2b+a^2c-abc)$$ or
$$\sum_{cyc}(a^5+a^4b+a^4c-a^3b^2-a^3c^2-6a^3bc+5a^2b^2c)\geq0.$$
Since by Schur $\sum\limits_{cyc}(a^5-a^4b-a^4c+a^3bc)\geq0$, it remains to prove that
$$\sum_{cyc}(2a^4b+2a^4c-a^3b^2-a^3c^2-7a^3bc+5a^2b^2c)\geq0$$ or
$$2\sum_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2)+\sum_{cyc}(a^3b^2+a^3c^2-2a^3bc)-5abc\sum_{cyc}(a^2-bc)\geq0$$ or
$$\sum_{cyc}(a-b)^2\left(2ab(a+b)+c^3-\frac{5}{2}abc\right)\geq0,$$ 
which is true by AM-GM:
$$2ab(a+b)+c^3-\frac{5}{2}abc\geq4\sqrt{a^3b^3}+c^3-\frac{5}{2}abc=2\cdot2\sqrt{a^3b^3}+c^3-\frac{5}{2}abc\geq$$
$$\geq3\sqrt[3]{4a^3b^3c^3}-\frac{5}{2}abc=\left(3\sqrt[3]4-\frac{5}{2}\right)abc>0.$$
Done!
