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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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Solution Half a solution | Comment 3 of 18 |
If any of the numbers is greater or equal than 1, the theorem holds. Say x≥1; then (x+y)^z>1^<>1 and (x+z)^y is also >1.

The problem to be analyzed is when x, y and z are all < 1... that's for later analysis!


  Posted by Federico Kereki on 2005-03-16 01:35:33
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