How do you prove the inequality
\begin{equation*} |\sqrt[3]{x} - \sqrt[3]{y}| \leq \sqrt[3]{|x-y|}? \end{equation*}
How do you prove the inequality
\begin{equation*} |\sqrt[3]{x} - \sqrt[3]{y}| \leq \sqrt[3]{|x-y|}? \end{equation*}
It's not true. Try $x=1$ and $y=-1$. If you assume $x$ and $y$ have the same sign, and you might as well assume $x\gt y\gt 0$, then it reduces to showing $(x-y)^3\leq x^3-y^3$ (WLOG replacing variables in the original inequality with cubes and cubing both sides). This is true because $x\gt y\gt 0$ implies $3xy^2<3x^2y$.