Absolute value inequality proof: $|a-b|^p\leq |a^p-b^p|$

Question

I have problem with this one inequality. How to solve it?

$|a-b|^p\leq |a^p-b^p|$,

$a,b\geq0$, $p\geq1$

I came up with and idea like

$|a-b|*|a-b|^{p-1}\leq |(a-b)*k|$

k is always $\geq0$, because

for p=2 right side looks like

$|(a-b)*(a+b)|$,

for p=3

$|(a-b)*(a^2+ab+b^2)|$

etc.

Then

$|a-b|^{p-1}\leq |k|.$

I don't know what to do next and I'm not sure, if that solve my problem.

Answer 1

First, notice that for every $x,y \geq 0$, one has $(x+y)^p \geq x^p + y^p$.

Let's assume that $0 \leq b \leq a$ (the other case being similar). Then the previous inequality applied with $x=b$ and $y=a-b$ gives $a^p \geq b^p + (a-b)^p$, so $(a-b)^p \leq a^p-b^p$, so $|a-b|^p \leq |a^p-b^p|$ since $a \geq b$.