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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 Second Half of Solution Comment 18 of 18 |
(In reply to Half a solution by Federico Kereki)

Note: Link to posted solution is not working.

Without loss of generality, let x>y>z

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

> (x+x)^x+(x+x)^x+(x+x)^x = 3*(2x)^x

Prove (2x)^x>2/3

Differentiating and setting = 0 yields x = (e^-1) / 2

and (2x)^x > 0.832 > 2/3

QED


  Posted by Andre on 2010-05-25 14:09:47
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