Let $k$ be the smallest positive real number such that for all positive real numbers $x$, we have

$$
\sqrt[3]{x} \leq k(x+1)
$$

What is the value of $k^{3}$?
We think in reverse by looking at the exponents of the left-hand side to guess how to dissect the right-hand side. The term $\sqrt[3]{x}$ must have been obtained from an AM-GM inequality with three terms. We could write $x+1=\frac{x}{2}+\frac{x}{2}+1$, but the AM-GM would then lead to a degree 2 on $x$. The correct dissection is therefore $x+1=x+\frac{1}{2}+\frac{1}{2}$, and it follows that

$$
x+1=x+\frac{1}{2}+\frac{1}{2} \geq 3 \sqrt[3]{x \times \frac{1}{2} \times \frac{1}{2}}
$$

which is

$$
(x+1) \frac{\sqrt[3]{4}}{3} \geq \sqrt[3]{x}
$$

Thus, $k \geq \frac{\sqrt[3]{4}}{3}$. To show that the constant $\frac{\sqrt[3]{4}}{3}$ is minimal, we need to find a value of $x$ for which there is equality in the inequality found. The equality case of the AM-GM suggests taking $x=\frac{1}{2}$. In conclusion, $k^{3}=\frac{4}{27}$.