Prove that $\sqrt[3]{a} - \sqrt[3]{b} < \sqrt[3]{a-b}$ by use of the MVT

Question

If we assume $a > b > 0$, how can the Mean Value Theorem prove this to be true?

Answer 1

For $x>b$ let

$f(x):=\sqrt[3]{x} - \sqrt[3]{b} - \sqrt[3]{x-b}$. Then we have $f(b)=0$.

If $a>b$ then there is $t \in (b,a)$ such that

$\frac{f(a)}{a-b}=f'(t)$

Its your turn to show that $f'(t)<0$