How can one prove that $\sqrt[3]{\left ( \frac{a^4+b^4}{a+b} \right )^{a+b}} \geq a^ab^b$, $a,b\in\mathbb{N^{*}}$? How can one prove that $\sqrt[3]{\left ( \frac{a^4+b^4}{a+b} \right )^{a+b}} \geq a^ab^b$, $a,b\in\mathbb{N^{*}}$?

This is expanding on a comment by Bill, the following might work:
You need 
$$ (a+b)\ln\sqrt[3]{(\frac{a^4+b^4}{a+b})} \geq a \ln(a) + b \ln(b)  \,.$$
Or
$$ (\ln\sqrt[3]{(\frac{a^4+b^4}{a+b})} \geq \frac{a}{a+b} \ln(a) + \frac{b}{a+b} \ln(b)  \,.$$
Now, if I remember right, the Jensen inequality for Log reads:
$$\frac{a}{a+b} \ln(a) + \frac{b}{a+b} \ln(b) \leq \ln (\frac{a^2+b^2}{a+b}) \,.$$
Thus, you only need to show 
$$\left( \frac{a^2+b^2}{a+b} \right)^3 \leq \frac{a^4+b^4}{a+b} \,.$$
Or
$$(a^2+b^2)^3 \leq (a+b)^2(a^4+b^4) \,.$$
EDIT
After a long calculation, this reduces to
$$a^6+3a^4b^2+3a^2b^4+b^6 \leq a^6+a^2b^4+2a^5b+2ab^5+a^2b^4+b^6$$
or
$$a^4b^2+a^2b^4 \leq a^5b+ab^5$$
After canceling $ab$ this follows imediatelly form the AM-GM.: $a^3b \leq \frac{a^4+a^4+a^4+b^4}{4}$ and $ab^3 \leq \frac{a^4+b^4+b^4+b^4}{4}$
