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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 $p$.

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1 Answer 1

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We can assume that $a \geq b$. For $x \in [0,1]$ the inequalities $$0 \leq 1-x^{1/p} \leq 1-x \leq (1 - x)^{1/p}$$ hold since $p>1$. Substituting $x \leftarrow b/a$ and raising to the power $p$ results in $$\left(1- \left(\frac{b}{a}\right)^{\frac{1}{p}}\right)^p \leq 1 - \frac{b}{a}$$ and after multiplication by $a$ $$\left(a^{\frac{1}{p}}- b^{\frac{1}{p}}\right)^p \leq a - b$$

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