How to prove the inequality $|x|^{r-1} \leq |x|^r + 1$

$|x|^{r-1} \leq |x|^r + 1$ of a convex function?

Thanks.

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What do you mean by a convex function here? –  Ilya Jan 9 '13 at 15:24
I mean $|x|^{p}$ is a convex function. –  RHS Jan 9 '13 at 15:25
Follow up question math.stackexchange.com/questions/274986/… –  RHS Jan 10 '13 at 7:18

First of all, if $r\lt1$, then multiplying the inequality by $|x|^{1-r}$ $$1\le|x|+|x|^{1-r}$$ which is false near $0$. So let's assume that $r\ge1$, then $$|x|^{r-1}\le1\quad\text{if }|x|\le1$$ and dividing both sides by $|x|^{r-1}$ yields $$|x|^{r-1}\le|x|^r\quad\text{if }|x|\ge1$$ Therefore, if $r\ge1$, $$|x|^{r-1}\le|x|^r+1$$

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It is hard to choose one, but I like this one with truncation better. –  RHS Jan 10 '13 at 6:47
First of all, for $r>0$ we have that $|x|^r\leq 1$ iff $|x|\leq 1$. Moreover, if $|x|>1$ then $$|x|^r = |x|\cdot|x|^{r-1}>|x|^{r-1}.$$ As a result, whenever $r-1>0$ (which is $r>1$) we have that $|x|^{r-1}\leq \max(1,|x|^r)\leq1+|x|^r$
@RHS This is a common method that comes without name. Ilya splits the domain and estimate the expression $|x|^r$ in each domain. Also, this has nothing to do with Jensen's inequality - it is just a crude estimate. –  AD. Jan 9 '13 at 15:34
I believe that $r\ge1$ is necessary. –  robjohn Jan 9 '13 at 19:07