<s>[INST] 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?
 [/INST] 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 $$ \frac{a^3}{b^2-bc+c^2}+
\frac{b^3}{c^2-ca+a^2} + \frac{c^3}{a^2-ab+b^2} \geq 3 \cdot
\frac{ab+bc+ca}{a+b+c}.$$
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