How to prove the Inequality $\sqrt[n]{b}-\sqrt[n]{a}<\sqrt[n]{b-a}$?

Question

Let $0<a<b$ using Lagrange mean value theorem I want to prove the following inequality

$\sqrt[n]{b}-\sqrt[n]{a}<\sqrt[n]{b-a}$

Any help is appreciated.

Answer 1

Consider $f(x) = \sqrt[n]{x-a}$. Since $b>a$, the interval $(a+a,b+a)$ is not empty and apply the Lagrange there. That is, we can find $\xi \in (2a, b+a)$ so that

$$ f(b+a) - f(2a) = f'(\xi) (b-a) $$

which reduces to

$$ \sqrt[n]{b} - \sqrt[n]{a} = \frac{1}{n} (\xi - a)^{1/n-1} (b-a) \leq \frac{1}{n}(b-a)^{1/n-1}(b-a) < \sqrt[n]{b-a} $$

Since $\xi \leq b$