For which real numbers does the following inequality hold?

$$
\sqrt[3]{5 x+2}-\sqrt[3]{x+3} \leq 1
$$
Let's introduce a new unknown; let $a=\sqrt[3]{x+3}$. Then the above inequality can be written as $\sqrt[3]{5\left(a^{3}-3\right)+1}-a \leq 1$. Rearranging this, we get:

$$
\sqrt[3]{5\left(a^{3}-3\right)+2} \leq a+1
$$

Cubing both sides and rearranging:

$$
\begin{gathered}
5\left(a^{3}-3\right)+2 \leq a^{3}+3 a^{2}+3 a+1 \\
4 a^{3}-3 a^{2}-3 a-14 \leq 0
\end{gathered}
$$

The solutions to this inequality are the same as those of the original, since cubing is an equivalent transformation. Factoring the expression on the left side of the inequality, we get:

$$
(a-2) \cdot\left(4 a^{2}+5 a+7\right) \leq 0
$$

The quadratic factor is always positive because its discriminant is negative. Thus, the product is non-positive if and only if $a-2 \leq 0$, which means $x \leq 5$.

Based on the work of Ernő Baranyi (Hódmezővásárhely, Bethlen G. Ref. Gymn., IV. grade)

Note. Several solved the problem by finding the equality instead of the inequality. However, the fact that equality holds at $x=5$ does not imply that the solution to the inequality is $x \leq 5$. Unfortunately, the function is not monotonic, so the reasoning of many was incorrect or incomplete. Many plotted the function, but few performed a complete function analysis, which was necessary for the proof.