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 $\sqrt[3]{\left ( \frac{a^4+b^4}{a+b} \right )^{a+b}} \geq a^ab^b$, $a,b\in\mathbb{N^{*}}$?
 A: 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.
A: 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 $ab^3 \leq \frac{a^4+b^4+b^4+b^4}{4}$
A: Here is a much more elementary proof:
$$a^{3a}b^{3b}=a^3a^3 \cdot... a^3 b^3b^3 \cdot ....b^3 \,.$$
Using the AM-GM inequality with $x_1=...=x_a=a^3$ and $x_{a+1}=...=x_{a+b}=b$ Yields
$$\sqrt[a+b]{a^3a^3 \cdot ... a^3 b^3b^3 \cdot ....b^3} \leq \frac{aa^3+bb^3}{a+b} \,.$$
Thus
$$a^{3a}b^{3b} \leq \left( \frac{a^4+b^4}{a+b} \right)^{a+b} \,.$$
A: With $p=a/(a+b)$ and $q=b/(a+b)$ you can use to homogeneity to get
$$\begin{align}
&&(a^4+b^4)^{a+b}&\ge (a+b)^{a+b}a^{3a}b^{3b} \\&\Leftrightarrow& (a^4+b^4)&\ge (a+b)a^{3p}b^{3q} \\ &\Leftrightarrow &p^4+q^4&\ge p^{3p}q^{3q}\\
& \Leftrightarrow  &
\sqrt[3]{p \cdot p^3 + q \cdot q^3} &\ge p^p q^q,
\end{align}$$ 
which is exactly the weighted mean inequality between the cubic mean and the geometric mean of $p$ and $q$ with weights $p$ and $q$.
(Obviously, the proofs of the general mean inequality are similar to the other posted answers, but one does not have to repeat the proof for each instance.)
