Let
$$S = \sqrt[3]{\frac{a}{b+7}} + \sqrt[3]{\frac{b}{c+7}} + \sqrt[3]{\frac{c}{d+7}} + \sqrt[3]{\frac{d}{a+7}}$$
Assume that $x$, $y$, $z$, $t$ is a permutation of the variables, with
$x \le y \le z \le t$. Then, by the rearrangement inequality,
$$S \le \left( \sqrt[3]{\frac{x}{t+7}} + \sqrt[3]{\frac{t}{x+7}} \right) + \left( \sqrt[3]{\frac{y}{z+7}} + \sqrt[3]{\frac{z}{y+7}} \right).$$

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

Notice that
$$\begin{aligned} 3\sqrt[3]{\frac{xt(x+t+14)}{7(x+7)(t+7)}} &= 3\sqrt[3]{\frac{t(x+7)}{7(t+7)} \cdot \frac{x(t+7)}{7(x+7)} \cdot \frac{7(x+t+14)}{(t+7)(x+7)}} \\ &\le \frac{t(x+7)}{7(t+7)} \cdot \frac{x(t+7)}{7(x+7)} \cdot \frac{7(x+t+14)}{(t+7)(x+7)} \end{aligned}$$
by the AM-GM inequality, so it suffices to prove
$$\frac{x}{t+7} + \frac{t}{x+7} + \frac{t(x+7)}{7(t+7)} + \frac{x(t+7)}{7(x+7)} + \frac{7(x+t+14)}{(t+7)(x+7)} \le \frac{x+t+14}{7}.$$
A straightforward check verifies that the last inequality is in fact an
equality. □

The claim leads now to
$$S \le \sqrt[3]{\frac{x+t+14}{7}} + \sqrt[3]{\frac{y+z+14}{7}} \le 2\sqrt[3]{\frac{x+y+z+t+28}{14}} = \frac{8}{\sqrt[3]{7}},$$
the last inequality being due to the AM-GM inequality (or to the fact
that $\sqrt[3]{}$ is concave on $[0, \infty)$).

Start by using Hölder's inequality:
$$\begin{aligned} S^3 &= \left( \sum_{cyc} \frac{\sqrt[6]{a} \cdot \sqrt[6]{a}}{\sqrt[3]{b+7}} \right)^3 \\ &\le \sum_{cyc} (\sqrt[6]{a})^3 \cdot \sum_{cyc} (\sqrt[6]{a})^3 \cdot \sum_{cyc} \left( \frac{1}{\sqrt[3]{b+7}} \right)^3 \\ &= \left( \sum_{cyc} \sqrt{a} \right)^2 \sum_{cyc} \frac{1}{b+7}. \end{aligned}$$

Notice that
$$\frac{(x-1)^2(x-7)^2}{x^2+7} \ge 0 \Leftrightarrow x^2 - 16x + 71 \ge \frac{448}{x^2+7}$$
yields
$$\begin{aligned} \sum \frac{1}{b+7} &\le \frac{1}{448} \sum (b - 16\sqrt{b} + 71) \\ &= \frac{1}{448} (384 - 16 \sum \sqrt{b}) \\ &= \frac{48 - 2 \sum \sqrt{b}}{56}. \end{aligned}$$
Finally,
$$\begin{aligned} S^3 &\le \frac{1}{56} \left( \sum \sqrt{a} \right)^2 \left( 48 - 2 \sum \sqrt{a} \right) \\ &\le \frac{1}{56} \left( \frac{\sum \sqrt{a} + \sum \sqrt{a} + (48 - 2\sqrt{a})}{3} \right)^3 \\ &= \frac{512}{7} \end{aligned}$$
by the AM-GM inequality. The conclusions follows.