Prove that for all positive x, y, and z,
(x+y)^z+(y+z)^x+(z+x)^y > 2.
(In reply to
re(3): Can get close to 2 - what is missing ? by pcbouhid)
Two or three of the variables being equal is not general enough. In all
of the instances you mention that involve small values or values
slightly less than 1, two or three of x, y, and z are always
equal. If you have pertinent modifications of the "e, 1-e"
results, you need to write them out in detail and show how they give a
proof for general values of x, y, and z no two of which are equal.
Consideration may however be limited to values of x, y, and z that are
between 0 and 1, since if any is 1 or greater, the result is clearly
true.
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Posted by Richard
on 2005-03-28 03:45:01 |
re(4): Can get close to 2 - what is missing ?
