Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Hy all! I'm having trouble finding a proof for the following problem:

Show that $a^x+b^x+c^x>(a+b+c)^x$, if $a,b,c>0$ and $0<x<1$ (over the real numbers).

This inequality has been torturing me for long hours the least. I couldn't find anything on Google. This kind of equation isn't made for search engines tbh.

Anyway, I really thinked about it a lot but didn't make any significant progress. It would be great to see a proof without the really high-end Mathematics. Any ideas to start with?

share|cite|improve this question
First divide out to reduce it to $a^x+b^x+c^x>1$, for $a, b, c<1$. Then use arithmetic-geometric inequality. – awllower Feb 10 '13 at 2:11
Essentially a duplicate of – sdcvvc Feb 10 '13 at 2:14
... because $(a+(b+c))^x\le a^x+(b+c)^x\le a^x+b^x+c^x$. – user53153 Feb 10 '13 at 3:14
up vote 1 down vote accepted

First, prove the inequality for two terms $a,b$. I essentially repeat the proof from Prove that $(p+q)^m \leq p^m+q^m$: let $y=1-x$ and
$$(a+b)^x = (a+b)^{1-y} = a (a+b)^{-y} + b (a+b)^{-y} < a a^{-y} + b b^{-y} = a^x + b^x$$ Now you can the inequality for three or more terms just by using parentheses:
$$(a+(b+c))^x < a^x+(b+c)^x < a^x+b^x+c^x$$ (This CW answer is a compilation of the comments above )

share|cite|improve this answer

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.