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

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

Set y=1-e, x=z=e where e is a small positive number. Then the expression evaluates to 2+(2*e)^(1-e)=2+2*e/(2*e)^e.  But (2*e)^e > M for some M>0 and for all e>0. (It can be shown using calculus that M ~= .83.) Thus the expression can be made less than 2+2*e/M for any e>0, that is, it can be made arbitrarily close to 2.

Posted by Richard on 2005-03-16 20:47:26