Inequality $\sum_{cyc}\frac{(a+b)^{3}}{\sqrt[3]{2(a+b)(a^{2}+b^{2})}}\ge{12}$ when $ab+bc+ca=6$ Let $a$, $b$ and $c$ be non-negative reals such that $ab+bc+ca=6$.
Prove that:
$$\frac{(a+b)^{3}}{\sqrt[3]{2(a+b)(a^{2}+b^{2})}}+\frac{(b+c)^{3}}{\sqrt[3]{2(b+c)(b^{2}+c^{2})}}+\frac{(c+a)^{3}}{\sqrt[3]{2(c+a)(c^{2}+a^{2})}}\ge{12}$$

Applying AM-GM on $2ab$ and $a^2+b^2$ we find $2ab(a^2+b^2) \leq \left(\frac{(a+b)^2}{2} \right)^2$ or $a^2+b^2 \leq \frac{(a+b)^4}{8ab}$. Plugging this in, we find
$$
\frac{(a+b)^3}{\sqrt[3]{2(a+b)(a^2+b^2)}} \geq \frac{(a+b)^3}{\sqrt[3]{2(a+b)\frac{(a+b)^4}{8ab}}} = \frac{(a+b)^3\sqrt[3]{4ab}}{\sqrt[3]{(a+b)^5}} = 2^{2/3} (a+b)^{4/3} (ab)^{1/3}. 
$$
Applying AM-GM on $a$ and $b$ we find $a+b \geq 2\sqrt{ab}$. Hence 
$$
\frac{(a+b)^3}{\sqrt[3]{2(a+b)(a^2+b^2)}} \geq 2^{2/3} (a+b)^{4/3} (ab)^{1/3} \geq 2^{2/3} (2\sqrt{ab})^{4/3} (ab)^{1/3} = 4ab.
$$
It follows that the given expression is at least $4(ab+bc+ca) = 24 \geq 12$.
