Does this hold for three numbers If $a\ge b\ge c\ge0$, does it hold that $\sqrt[3]{\left(a-b+c\right)^{2}}\ge\sqrt[3]{a^{2}}-\sqrt[3]{b^{2}}+\sqrt[3]{c^{2}}$?
Thanks for any help.

Consider one case of $a = b \ge  c \ge 0$. Both sides are equal, and the inequality holds.
Now, partial differentiate $\sqrt[3]{(a-b+c)^2} - \sqrt[3]{a^2} + \sqrt[3]{b^2} - \sqrt[3]{c^2}$ with respect to a.
$$\frac{2}{3}(a-b+c)^{-\frac{1}{3}} - \frac{2}{3}a^{-\frac{1}{3}}$$
$$=\frac{2}{3\sqrt[3]{a-b+c}} - \frac{2}{3\sqrt[3]{a}}$$
Show this partial derivative is positive (because $-b+c$ is negative), and you are done.
