I am trying to prove that for any $p > 1$ and for any real numbers $a,b > 0$, the following inequality holds: $ | a^{\frac{1}{p}} - b^{\frac{1}{p}} |^p \leq 2^p|a-b| $
In the case where $p$ is a positive integer, it seems like I could construct an even tighter bound than the one given: $ | a^{\frac{1}{p}} - b^{\frac{1}{p}} | + \min(a^{\frac{1}{p}},b^{\frac{1}{p}}) = \max(a^{\frac{1}{p}},b^{\frac{1}{p}}) $ thus raising both sides to the $p$-th power, applying the binomial theorem and noting that all the cross-terms in the binomial expansion are nonnegative, we have $ | a^{\frac{1}{p}} - b^{\frac{1}{p}} |^p + \min(a^{\frac{1}{p}},b^{\frac{1}{p}})^p \leq \max(a^{\frac{1}{p}},b^{\frac{1}{p}})^p \\ \implies | a^{\frac{1}{p}} - b^{\frac{1}{p}} |^p \leq \max(a,b) - \min(a,b) = |a-b| \leq 2^p |a-b| $
However, I am not seeing how to generalize this argument to non-integer $p$.