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Power inequality (Posted on 2005-03-15) Difficulty: 3 of 5
Prove that for all positive x, y, and z,

(x+y)^z+(y+z)^x+(z+x)^y > 2.

See The Solution Submitted by e.g.    
Rating: 3.0000 (1 votes)

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Some Thoughts re(3): am I wrong ? | Comment 5 of 18 |
(In reply to re(2): am I wrong ? by pcbouhid)

I found that when x=y=z~0.183935 that the minimum value is approximately 2.495957862. I believe this is the low value for x=y=z.

I don't think there is a lower value when x, y and z are not equal, but my proof doesn't exist.
  Posted by Michael Cottle on 2005-03-16 17:57:18

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