Prove an inequality concerning $\sqrt[3]{4a^3+4b^3}+\sqrt[3]{4b^3+4c^3}+\sqrt[3]{4c^3+4a^3}$ Let $a,b,c$ be positive. I need to prove
$\sqrt[3]{4a^3+4b^3}+\sqrt[3]{4b^3+4c^3}+\sqrt[3]{4c^3+4a^3}\leq \dfrac{4a^2}{a+b}+\dfrac{4b^2}{b+c}+\dfrac{4c^2}{c+a}$
Thanks!

Observe that 
$$\sum \frac{4a^2 - 4b^2} { a+b} = \sum (4a-4b) = 0. $$
Hence, we have
$$\sum \frac{4a^2} { a+b}  = \sum \frac{4b^2} {a+b}$$
Hence, we may rewrite the RHS and make it more symmetric, namely
$$\sum \sqrt[3]{4a^3+4b^3} \leq \sum \frac{4a^2} {a+b} = \sum \frac{ 2a^2+2b^2} { a+b}.$$
Now, we can break it up into the individual terms, and just show that
$$ \frac{ a^3+b^3} {2} \left( \frac{a+b}{2} \right)^3 \leq \left( \frac{a^2+b^2} {2} \right)^3$$
This is true because

 [Newton] $$S_1 S_3 \leq S_2^2$$
 [Maclaurin] $$S_1^2 \leq S_2$$

