# How to prove that $|x|^{r-1} \leq |x|^r + 1$

I'm trying to show that $|x|^{r-1} \leq |x|^r + 1$ with the help of Jensen's inequality? Thanks.

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I am only looking for solution using Jensen's inequality here. Thanks. – RHS Jan 10 '13 at 7:23
There exists much simpler proofs. Why insisting on using Jensen? – Did Jan 10 '13 at 7:45
Because I think the one involved with cases isn't too beautiful... – RHS Jan 10 '13 at 7:57
OK. And why do you think a (not too contrived) proof using Jensen is simply possible? There is no structure à la Jensen here (for example the equality is never realized). – Did Jan 10 '13 at 8:10
Because I noticed $|x|^p$ is a convex function in first glance. And the above inequality's form look very Jensen's inequality. Sorry what did you mean by the equality is never realized? – RHS Jan 10 '13 at 8:15

WLOG, we may assume that $x>0.$ Thus, $x=e^z$ for some $z.$ This gives $$e^{z(r-1)} \le e^{zr}+1$$ and it should now be a bit easier to apply Jensen's inequality to the convex function $e^t$.
Then apply $z=lg|y|$? Do you have assumption $z>0$ here? – RHS Jan 10 '13 at 8:34
RHS: To apply $x=e^z$ and then $z=\log|y|$ might be described as running in circles. – Did Jan 10 '13 at 8:41
Why? Isn't it will then yield the form above? For $z > 0$ we have the above inequality by Paxinum. Then put $z = lg|y|$, where $|y| \geq 1$. We have, $e^{lg|y|(r-1)} \leq e^{in|y|r} + 1$. And finially $|y|^{r-1} \leq |y|^r + 1$. – RHS Jan 10 '13 at 10:31