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}

AM-GM inequality

A very useful inequality in Mathematics is the AM-GM Inequality.

The arithmetic mean of numbers $x_1, x_2, \cdots, x_n$ is $\displaystyle \boxed{\frac{x_1+ x_2+\cdots+x_n}{n}}$.

The geometric mean of numbers $x_1, x_2, \cdots, x_n$ is $\boxed{\sqrt[n]{x_1\cdot x_2 \cdots x_n}}$.

The AM-GM Inequality states that:

For any nonnegative numbers $x_1, x_2, \cdots, x_n$,

$\displaystyle\boxed{\frac{x_1+x_2+\cdots+ x_n}{n}\geq\sqrt[n]{x_1\cdot x_2 \cdots x_n}}$, and equality holds if and only if $x_1=x_2=\cdots=x_n$.

How to Apply?

Let say we have three (nonnegative) numbers a, b, c that add up to 30, i.e. $a+b+c=30$. Can we know what is the largest possible product $abc$?

Yes! Using the AM-GM inequality we have just learnt above, we know $\displaystyle \frac{a+b+c}{3}\geq \sqrt[3]{abc}$.

$\displaystyle 10\geq \sqrt[3]{abc}$

Cubing both sides, we have, $\displaystyle abc\leq 10^3=1000$.

Also, the AM-GM inequality tells us that there is equality only when $a=b=c$, i.e. $a=b=c=10$. Hence, the largest possible product $abc$ is 1000.

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