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}

AM-GM inequality

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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