Inequalities for pth powers, where 0<p<infinity

There are some useful inequalities for |x+y|^p, where p is a number ranging from 0 to infinity. These are the top 3 useful inequalities (note some of them only work for p less than 1, or p greater than 1).

For a,b\in\mathbb{R}, |a+b|^p\leq 2^p(|a|^p+|b|^p), where 0<p<\infty.

\begin{aligned}  |a+b|^p&\leq(|a|+|b|)^p\\  &\leq(2\max\{|a|,|b|\})^p\\  &=2^p(\max\{|a|,|b|\})^p\\  &\leq 2^p(|a|^p+|b|^p).  \end{aligned}

If 0<p<1, |a+b|^p\leq|a|^p+|b|^p for all a,b\in\mathbb{R}.

\displaystyle 1=\frac{|a|}{|a|+|b|}+\frac{|b|}{|a|+|b|}\leq\left(\frac{|a|}{|a|+|b|}\right)^p+\left(\frac{|b|}{|a|+|b|}\right)^p=\frac{|a|^p+|b|^p}{(|a|+|b|)^p}.
Hence |a+b|^p\leq(|a|+|b|)^p\leq|a|^p+|b|^p.

For a,b\in\mathbb{R}, |a+b|^p\leq 2^{p-1}(|a|^p+|b|^p) for 1\leq p<\infty.

By convexity of |x|^p for 1\leq p<\infty, \displaystyle \left|\frac 12 a+\frac 12 b\right|^p\leq\frac 12 |a|^p+\frac 12 |b|^p.
Multiplying throughout by 2^p gives \displaystyle |a+b|^p\leq 2^{p-1}(|a|^p+|b|^p).

Author: mathtuition88

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