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

Since $\log(x)$ is concave,
$$
\log\left(\frac{ax+by}{a+b}\right)\ge\frac{a\log(x)+b\log(y)}{a+b}\tag{1}
$$
Rearranging $(1)$ and exponentiating yields
$$
\left(\frac{ax+by}{a+b}\right)^{a+b}\ge x^ay^b\tag{2}
$$
Plugging $x=a^3$ and $y=b^3$ into $(2)$ gives
$$
\left(\frac{a^4+b^4}{a+b}\right)^{a+b}\ge a^{3a}b^{3b}\tag{3}
$$
and $(3)$ is the cube of the posited inequality.
From my comment (not using concavity):
For $0<t<1$, the minimum of $t+(1-t)u-u^{1-t}$ occurs when $(1-t)-(1-t)u^{-t}=0$; that is, when $u=1$. Therefore, $t+(1-t)u-u^{1-t}\ge0$. If we set $u=\frac{y}{x}$ and $t=\frac{a}{a+b}$, we get
$$
\frac{ax+by}{a+b}\ge x^{a/(a+b)}y^{b/(a+b)}\tag{4}
$$
Inequality $(2)$ is simply $(4)$ raised to the $a+b$ power.
