x^y + y^x <1

From: Amanda (sca18_at_hotmail.com)
Date: 12/27/04 

Date: 27 Dec 2004 07:37:44 -0800

Hello

I'd like some hints to prove that, if x,y>0 then x^y + y^x >1. If we
keep y fixed and define f(x) = x^y + y^x, then, if y>=1, x^y and y^x
are increasing functions of x, so that f(x) > lim (x->0) f(x) =1. In
virtue of the symmetry of the function, we see that the inequality
holds whenever x>=1 or y>=1.
But if x and y are in (0,1), I got confused. We have f'(x) =
y*(x^(y-1)) + (y^x)*ln(y). f'(x) -> oo when x->0+, so f is stricly
increasing on (0,a) for some a<1. we have f'(1) = y + y*ln(y), so f'
will have at least one root in (0,1) if y <= 1/e. I think f' has no
root in (0,1) if y >1/e, but I'm not sure.
Any hint is welcome.
Thank you.
Amanda