Re: x^y + y^z + z^x > 1

On Sat, 01 Dec 2007 12:16:41 +0000, I wrote:

On Fri, 30 Nov 2007 23:32:34 +0000, I wrote:

(My solution was very similar, just somewhat longer and messier.)

For what it's worth, the slight difference was that in the last
part [...]

Oh, and I made the first part unnecessarily complicated too, by
arguing that w.l.o.g., by symmetry, either 0 < x <= y <= z < 1
or 0 < z <= y <= x < 1, after either of which there is a pretty
random choice of how to use the result x^y + y^x > 1 (with one
or another substitution), which is also "messy" (albeit simple).
--
Angus Rodgers
(twirlip@ eats spam; reply to angusrod@)
Contains mild peril
.

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