Show $(|a|+|b|)^p \leq |a|^p+|b|^p$ for $0\leq p \leq 1$

Question

How to show that for any real numbers $a$ and $b$ $$(|a|+|b|)^p \leq |a|^p+|b|^p$$ for $0\leq p \leq 1$ ?

Answer 1

The inequality is equivalent to $$ x^p+y^p\ge 1, \text{ for }x+y=1,x,y\ge0.$$ Noting that, for $p\in[0,1],x,y\in[0,1]$, $$ x^p\ge x,y^p\ge y $$ one has $$ x^p+y^p\ge 1. $$

Answer 2

You want $(a+b)^p \le a^p+b^p $ with $a \ge 0, b \ge 0, 0 \le p \le 1$.

We can assume that $a \le b $.

If $a=0$ this is $b^p \le b^p$, which is true. So we can assume that $a > 0$.

If $a=b$, this becomes $2^pa^p \le 2a^p $ or $2^p \le 2$ which is true. So we can assume $0 < a < b$.

Dividing by $b^p$, this is $(r+1)^p \le r^p+1 $ where $r = \frac{a}{b} $ and $0 < r < 1$.

If $p=1$ this is $r+1 \le r+1$ which is true, so we can assume that $0 \le p < 1$.

Let $f(r) = r^p + 1 - (r+1)^p $.

$f(0) =0 $, $f(1) =2-2^p \ge 0$, and

$\begin{array}\\ f'(r) &=pr^{p-1}-p(r+1)^{p-1}\\ &=pr^{p-1}(1-(1+1/r)^{p-1})\\ &\ge 0\\ \end{array} $

Since $1+1/r > 1$ and $-1 \le p-1 < 0$, $(1+1/r)^{p-1} \lt 1$, so $f'(r) \ge 0$ so $f(r) \ge 0$ which is what we want.

This is probably too long, but this is what I did.