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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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re(3): Can get close to 2 - what is missing ? | Comment 15 of 18 |
(In reply to re(2): Can get close to 2 - what is missing ? by Richard)

x, y and z are positive numbers.

It is proved that for small value of only one of them, for small values of two of them and for small values of all three, the expression evaluate to a number greater than 2.

Since in the two first cases (where it is used "1-e"), with the substitution of the "1" for a number greater than 1, say N (and now we use "N-e"), the expression evaluate to a number much greater than the values that we found in a), I don't know how could exist positive numbers that makes the expression less than or equal 2.

  


  Posted by pcbouhid on 2005-03-27 21:45:48
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