## 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 $|\frac{a+b}{2}|^p\leq\frac{1}{2}|a|^p+\frac{1}{2}|b|^p$.

Thus,

\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}