Is the inequality $| \sqrt[3]{x^2} - \sqrt[3]{y^2} | \le \sqrt[3]{|x -y|^2}$ true? I'm having some trouble deciding whether this inequality is true or not...
$| \sqrt[3]{x^2} - \sqrt[3]{y^2} | \le \sqrt[3]{|x -y|^2}$ for $x, y \in \mathbb{R}.$

It is shown in Is $x^t$ subadditive for $t \in [0,1]$? that we have the inequality
$$(a+b)^t \le a^t + b^t \tag{1}$$
for all nonnegative $a,b$ and all $t \in [0,1]$.
For your inequality, let us assume without loss of generality that $x \ge y$.  Applying the above inequality (1) with $a=y$, $b=x-y$, and $t=2/3$ we get
$$x^{2/3} \le y^{2/3} + (x-y)^{2/3}.$$
Subtracting $y^{2/3}$ from both sides,
$$x^{2/3} - y^{2/3} \le (x-y)^{2/3}.$$
Now since $x \ge y$ we have $x-y = |x-y|$.  And since the function $x \mapsto x^{2/3}$ is increasing, we have $x^{2/3} \ge y^{2/3}$, so $x^{2/3} - y^{2/3} = |x^{2/3} - y^{2/3}|$.  Thus we showed
$$|x^{2/3} - y^{2/3}| \le |x-y|^{2/3}$$
which is the desired inequality.
