Curious Inequality: 2^p(|a|^p+|b|^p)>=|a+b|^p

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This inequality often appears in Analysis: 2^p(|a|^p+|b|^p)\geq |a+b|^p, for p\geq 1, and a,b\in\mathbb{R}. It does seem quite tricky to prove, and using Binomial Theorem leads to a mess and doesn’t work!

It turns out that the key is to use convexity, and we can even prove a stronger version of the above, namely 2^{p-1}(|a|^p+|b|^p)\geq |a+b|^p.

Proof: Consider f(x)=|x|^p which is convex on \mathbb{R}. Let a,b\in\mathbb{R}. By convexity, we have \displaystyle\boxed{f(ta+(1-t)b)\leq tf(a)+(1-t)f(b)} for 0\leq t\leq 1.

Choose t=1/2. Then we have f(\frac{1}{2}a +\frac 12 b)\leq \frac 12 f(a)+\frac 12 f(b), which implies |\frac{a+b}{2}|^p\leq\frac{1}{2}|a|^p+\frac{1}{2}|b|^p.


\begin{aligned}|a+b|^p&\leq 2^{p-1}|a|^p+2^{p-1}|b|^p\\    &\leq 2^p|a|^p+2^p|b|^p\\    &=2^p(|a|^p+|b|^p)    \end{aligned}

Author: mathtuition88

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