Prove that for all positive x, y, and z,
(x+y)^z+(y+z)^x+(z+x)^y > 2.
(In reply to
Can get close to 2 by Richard)
Sorry, Richard, for the my insistence, or for, perhaps, my misunderstanding.
As you showed, seting y=(1-e), x=z=e, where e is a small positive number, the expression evaluates to 2 + (2*e)^(1-e). (2*e)^e effectivelly has an inflexion point in .83, but (2*e)^(1-e) tends to zero. So, the whole expression can be made, as you said, arbitrarily, close to 2, but not less than 2.
If you set x=y=(1-e) and z =e, the expression evaluates to 2 + ((2/(1-e))^e, and the second term tends to 1, so the whole expression tends to 3.
If you set x=y=z=(1-e), the expression evaluates to 3 * ((2*(1-e))^(1-e), and the second term also tends to 1, so the whole expression tends, also, to 3.
Finally, if we set x=y=z=e, the expression evaluates to 3*(2*e)^e, and since (2*e)^e has an inflexion point in 0.83..., the whole expression tends to 2,49...
What is missing ?
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Posted by pcbouhid
on 2005-03-27 17:21:54 |
re: Can get close to 2 - what is missing ?
