<s>[INST] If $a\ge b\ge-c\ge0$, is $\sqrt[3]{a-b-c}\ge\sqrt[3]{a}-\sqrt[3]{b}-\sqrt[3]{c}$? Let $a\ge b\ge-c\ge0$. Is it true that $\sqrt[3]{a-b-c}\ge\sqrt[3]{a}-\sqrt[3]{b}-\sqrt[3]{c}$?
 [/INST] Sure.
Since $a\geq 0$ and $-c \geq 0$, $a-c\geq 0$. Therefore
$$(a-b-c)-(\sqrt[3]{a}-\sqrt[3]{b}-\sqrt[3]{c})^3 = 3(\sqrt[3]{a}-\sqrt[3]{c})(\sqrt[3]{b}-\sqrt[3]{-c})(\sqrt[3]{a}-\sqrt[3]{b}) \geq 0$$
since all of the factors on the right are non-negative.
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