Prove that for all positive x, y, and z,

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

Given that X and Y and Z are NOT equal (b/c they are not set equal) or negative, then you must give X, Y, and Z positive, ≠ values.  So, the proof the above equation works...
EXAMPLE:
1) Give X=1x10^-28, Y=1x10^-29, and Z=1x10^-30.
2) Plug these numbers into the equation like so:
([1x10^-28]+[1x10^-29])^(1x10^-30)+
([1x10^-29]+[1x10^-30])^(1x10^-28)+
([1x10^-30]+[1x10^-28])^(1x10^-29)
3)Do the math.
4)You should get the answer '3'.
In fact, I have found that if you do this with X=1x10^-32, Y=1x10^-33, and Z=1x10^-34, you will get the same answer.
Given these facts, it is hard to say that anything following the above rules will turn out below 2.

Posted by David on 2005-03-28 07:20:39