A7. Find the maximal value of
$$
S=\sqrt[3]{\frac{a}{b+7}}+\sqrt[3]{\frac{b}{c+7}}+\sqrt[3]{\frac{c}{d+7}}+\sqrt[3]{\frac{d}{a+7}},
$$
where $a, b, c, d$ are nonnegative real numbers which satisfy $a+b+c+d=100$.
Answer: $\frac{8}{\sqrt[3]{7}}$, reached when $(a, b, c, d)$ is a cyclic permutation of $(1,49,1,49)$.
Solution 1. Since the value $8 / \sqrt[3]{7}$ is reached, it suffices to prove that $S \leqslant 8 / \sqrt[3]{7}$.
Assume that $x, y, z, t$ is a permutation of the variables, with $x \leqslant y \leqslant z \leqslant t$. Then, by the rearrangement inequality,
$$
S \leqslant\left(\sqrt[3]{\frac{x}{t+7}}+\sqrt[3]{\frac{t}{x+7}}\right)+\left(\sqrt[3]{\frac{y}{z+7}}+\sqrt[3]{\frac{z}{y+7}}\right) .
$$

Claim. The first bracket above does not exceed $\sqrt[3]{\frac{x+t+14}{7}}$.
Proof. Since
$$
X^{3}+Y^{3}+3 X Y Z-Z^{3}=\frac{1}{2}(X+Y-Z)\left((X-Y)^{2}+(X+Z)^{2}+(Y+Z)^{2}\right),
$$
the inequality $X+Y \leqslant Z$ is equivalent (when $X, Y, Z \geqslant 0$ ) to $X^{3}+Y^{3}+3 X Y Z \leqslant Z^{3}$. Therefore, the claim is equivalent to
$$
\frac{x}{t+7}+\frac{t}{x+7}+3 \sqrt[3]{\frac{x t(x+t+14)}{7(x+7)(t+7)}} \leqslant \frac{x+t+14}{7} .
$$

Notice that
$$
\begin{array}{r}
3 \sqrt[3]{\frac{x t(x+t+14)}{7(x+7)(t+7)}}=3 \sqrt[3]{\frac{t(x+7)}{7(t+7)} \cdot \frac{x(t+7)}{7(x+7)} \cdot \frac{7(x+t+14)}{(t+7)(x+7)}} \\
\leqslant \frac{t(x+7)}{7(t+7)}+\frac{x(t+7)}{7(x+7)}+\frac{7(x+t+14)}{(t+7)(x+7)}
\end{array}
$$
by the AM-GM inequality, so it suffices to prove
$$
\frac{x}{t+7}+\frac{t}{x+7}+\frac{t(x+7)}{7(t+7)}+\frac{x(t+7)}{7(x+7)}+\frac{7(x+t+14)}{(t+7)(x+7)} \leqslant \frac{x+t+14}{7} .
$$

A straightforward check verifies that the last inequality is in fact an equality.
The claim leads now to
$$
S \leqslant \sqrt[3]{\frac{x+t+14}{7}}+\sqrt[3]{\frac{y+z+14}{7}} \leqslant 2 \sqrt[3]{\frac{x+y+z+t+28}{14}}=\frac{8}{\sqrt[3]{7}},
$$
the last inequality being due to the AM-CM inequality (or to the fact that $\sqrt[3]{ }$ is concave on $[0, \infty))$.
Solution 2. We present a different proof for the estimate $S \leqslant 8 / \sqrt[3]{7}$.
Start by using Hölder's inequality:
$$
S^{3}=\left(\sum_{\text {cyc }} \frac{\sqrt[6]{a} \cdot \sqrt[6]{a}}{\sqrt[3]{b+7}}\right)^{3} \leqslant \sum_{\text {cyc }}(\sqrt[6]{a})^{3} \cdot \sum_{\text {cyc }}(\sqrt[6]{a})^{3} \cdot \sum_{\text {cyc }}\left(\frac{1}{\sqrt[3]{b+7}}\right)^{3}=\left(\sum_{\text {cyc }} \sqrt{a}\right)^{2} \sum_{\mathrm{cyc}} \frac{1}{b+7} .
$$

Notice that
$$
\frac{(x-1)^{2}(x-7)^{2}}{x^{2}+7} \geqslant 0 \Longleftrightarrow x^{2}-16 x+71 \geqslant \frac{448}{x^{2}+7}
$$
yields
$$
\sum \frac{1}{b+7} \leqslant \frac{1}{448} \sum(b-16 \sqrt{b}+71)=\frac{1}{448}\left(384-16 \sum \sqrt{b}\right)=\frac{48-2 \sum \sqrt{b}}{56} .
$$

Finally,
$$
S^{3} \leqslant \frac{1}{56}\left(\sum \sqrt{a}\right)^{2}\left(48-2 \sum \sqrt{a}\right) \leqslant \frac{1}{56}\left(\frac{\sum \sqrt{a}+\sum \sqrt{a}+\left(48-2 \sum \sqrt{a}\right)}{3}\right)^{3}=\frac{512}{7}
$$
by the AM-GM inequality. The conclusion follows.
Comment. All the above works if we replace 7 and 100 with $k>0$ and $2\left(k^{2}+1\right)$, respectively; in this case, the answer becomes
$$
2 \sqrt[3]{\frac{(k+1)^{2}}{k}}
$$

Even further, a linear substitution allows to extend the solutions to a version with 7 and 100 being replaced with arbitrary positive real numbers $p$ and $q$ satisfying $q \geqslant 4 p$.