3

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?

Ralph
  • 331
  • 2
  • 13
  • 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 02:11
  • 2
    Essentially a duplicate of http://math.stackexchange.com/questions/264156/prove-that-pqm-leq-pmqm – sdcvvc Feb 10 '13 at 02:14
  • 2
    ... because $(a+(b+c))^x\le a^x+(b+c)^x\le a^x+b^x+c^x$. –  Feb 10 '13 at 03:14

1 Answers1

1

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 )

Community
  • 1
ˈjuː.zɚ79365
  • 3,134