7. For non-negative real numbers $a, b, c, d$, if $a+b+c+d=100$, find
$$
S=\sqrt[3]{\frac{a}{b+7}}+\sqrt[3]{\frac{b}{c+7}}+\sqrt[3]{\frac{c}{d+7}}+\sqrt[3]{\frac{d}{a+7}}
$$

the maximum value.
7. The maximum value sought is $\frac{8}{\sqrt[3]{7}}$, which can be achieved when $(a, b, c, d)$ is $(1,49,1,49)$ and its cyclic permutations.
Since $\frac{8}{\sqrt[3]{7}}$ can be achieved, it suffices to prove that $S \leqslant \frac{8}{\sqrt[3]{7}}$.
Assume $x, y, z, t$ are a permutation of the variables, and $x \leqslant y \leqslant z \leqslant t$.
By the rearrangement inequality, we have
$$
S \leqslant\left(\sqrt[3]{\frac{x}{t+7}}+\sqrt[3]{\frac{t}{x+7}}\right)+\left(\sqrt[3]{\frac{y}{z+7}}+\sqrt[3]{\frac{z}{y+7}}\right) .
$$

First, we prove a lemma.
Lemma: $\sqrt[3]{\frac{x}{t+7}}+\sqrt[3]{\frac{t}{x+7}} \leqslant \sqrt[3]{\frac{x+t+14}{7}}$.
Proof: Because
$$
\begin{array}{l} 
X^{3}+Y^{3}+3 X Y Z-Z^{3} \\
= \frac{1}{2}(X+Y-Z)\left((X-Y)^{2}+(X+Z)^{2}+\right. \\
\left.(Y+Z)^{2}\right),
\end{array}
$$

for $X, Y, Z \geqslant 0$,
$$
X+Y \leqslant Z \Leftrightarrow X^{3}+Y^{3}+3 X Y Z \leqslant Z^{3} \text {. }
$$

Thus, inequality (1) is equivalent to
$$
\begin{array}{l}
\frac{x}{t+7}+\frac{t}{x+7}+3 \sqrt[3]{\frac{x t(x+t+14)}{7(x+7)(t+7)}} \\
\leqslant \frac{x+t+14}{7} .
\end{array}
$$

By the AM-GM inequality, we have
$$
\begin{array}{l}
3 \sqrt[3]{\frac{x t(x+t+14)}{7(x+7)(t+7)}} \\
=3 \sqrt[3]{\frac{t(x+7)}{7(t+7)} \cdot \frac{x(t+7)}{7(x+7)} \cdot \frac{7(x+t+14)}{(t+7)(x+7)}} \\
\leqslant \frac{t(x+7)}{7(t+7)}+\frac{x(t+7)}{7(x+7)}+\frac{7(x+t+14)}{(t+7)(x+7)} .
\end{array}
$$

Therefore, to prove inequality (2), it suffices to prove
$$
\begin{array}{l}
\frac{x}{t+7}+\frac{t}{x+7}+\frac{t(x+7)}{7(t+7)}+ \\
\frac{x(t+7)}{7(x+7)}+\frac{7(x+t+14)}{(t+7)(x+7)} \\
\leqslant \frac{x+t+14}{7} .
\end{array}
$$

Direct verification shows that equality holds in inequality (3).
The lemma is proved.
By the lemma, we have
$$
\begin{array}{l}
S \leqslant \sqrt[3]{\frac{x+t+14}{7}}+\sqrt[3]{\frac{y+z+14}{7}} \\
\leqslant 2 \sqrt[3]{\frac{x+y+z+t+28}{14}}=\frac{8}{\sqrt[3]{7}} .
\end{array}
$$

The last inequality uses Jensen's inequality, since $f(u)=\sqrt[3]{u}$ is a concave function on the interval $[0,+\infty)$.