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Let $a$ and $b$ be real numbers and $p>0$. What is the best way to prove that $|a+b|^p<2^p(|a|^p+|b|^p)$?

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4 Answers 4

up vote 2 down vote accepted

If you divide $2^p$ on both sides, then you get $(|a+b|/2)^p < |a|^p + |b|^p$. Can you see why that would be true?

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lol, that's really very easy –  Jacques Nov 20 '10 at 18:59

Well,

\begin{align*} |a + b|^p \leq (|a| + |b|)^p &\leq 2^p \text{max}\{|a|^p, |b|^p\}\\ &\leq 2^{p - 1} (|a|^p + |b|^p - |a^p - b^p|)\\ &\leq 2^{p - 1} (|a|^p + |b|^p) \end{align*}

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+1 For the stars!! :) –  AD. Nov 20 '10 at 21:53

If $0<p\leq1$ you don't need any powers of $2$. This came up in another recent question. If $1\leq p$ you can strengthen the inequality to $|a+b|^p\leq 2^{p-1}(|a|^p+|b|^p)$ by applying convexity of the function $t\mapsto t^p$.

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Here is an alternative for the case $p\ge1$ we may also use Hölder's inequality $$|a+b|=|1\cdot a +1\cdot b|\leq (1^q+ 1^q)^{1/q}(|a|^p+|b|^p)^{1/p}=2^{1/q}(|a|^p+|b|^p)^{1/p}$$ where $1/q+1/p=1$, which reduces to $$|a+b|^p\le 2^{p/q}(|a|^p+|b|^p)=2^{p(1-1/p)}(|a|^p+|b|^p)=2^{p-1}(|a|^p+|b|^p).$$ For $0<p<1$, please see my answer in the thread "Please explain inequality $|x^p−y^p|\le|x−y|^p$"

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