Prove that $|a+b|^p \leq 2^p \{ |a|^p +|b|^p \}$

For any real numbers $a$ and $b$ and $1 \leq p < \infty$, prove that

$$|a+b|^p \leq 2^p \{ |a|^p +|b|^p \}$$

This inequality is given in the the book Real Analysis by Royden, Chapter $7$, page $136$. I don't understand how the author comes to this inequality. Can anyone provide some hints?

Solution 1:

It can be quite simple: let $c=\max\{|a|,|b|\}$ so that $$ |a+b|\leq |a|+|b|\leq 2c\implies |a+b|^p\leq (2c)^p=2^p(c^p)\leq 2^p(|a|^p+|b|^p). $$ In fact, you only need $p\geq 0$.

Edit: As Winther indicates elsewhere in this thread, $p\geq 1$ gives you a stronger result. I'll give a proof using the convexity of a different function than one he suggests. Let $g(x)=x^p$ defined on $[0,\infty)$ and $m=|a|$ and $n=|b|$. We would like to prove $|a+b|^p\leq 2^{p-1}(|a|^p+|b|^p)$. By the triangle inequality above, it suffices to show \begin{align*} (m+n)^p\leq 2^{p-1}(m^p+n^p)&\iff\left(\frac{1}{2}m+\frac{1}{2}n\right)^p\leq\frac{1}{2}m^p+\frac{1}{2}n^p\\ &\iff g\left(\frac{1}{2}m+\frac{1}{2}n\right)\leq\frac{1}{2}g(m)+\frac{1}{2}g(n) \end{align*} which is true due to the convexity of $g$. It is here that we need $p\geq 1$.

Solution 2:

Hint:

A convex function always satisfy $$f(tx+(1-t)y)\leq tf(x) + (1-t)f(y),~~~~~~~t\in[0,1]$$

Take $f(x) = |x|^p$, show that this is convex for $p\geq 1$ (for example by the second derivative test) and find a suitable value for $t$.

This gives the slightly better inequality: $|a+b|^p \leq 2^{p-1}(|a|^p + |b|^p)$.