Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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 $e^t$.

share|cite|improve this answer
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
RHS: You are doing a circular argument. I rewrote your unproven inequality, to an equivalent, unproven inequality. You need to prove my inequality with Jensen's. – Per Alexandersson Jan 10 '13 at 17:24
But how to get my inequality after proving yours? – RHS Jan 18 '13 at 13:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.