Re: x^y + y^x <1
From: Todd Trimble (trimble1_at_optonline.net)
Date: 12/28/04
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Date: Tue, 28 Dec 2004 02:47:44 +0000 (UTC)
On 27 Dec 2004, Amanda wrote:
>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
Let's restrict to (0, 1) x (0, 1) and suppose WLOG that x > y
(the case x = y is easily disposed of). Then either x >= 1/e
or y < 1/e.
Case 1: x >= 1/e. Then x^y + y^x (for x fixed) increases
as a function of y in the domain 0 < y < x, since
x y^(x-1) + x^y ln(x) >= x y^(x-1) - x^y > 0
using the lemma below. Hence... (consider y -> 0).
Case 2: y < 1/e. Then x^y + y^x (for y fixed) decreases
as a function of x in the domain y < x < 1, since
y x^(y-1) + y^x ln(y) < y x^(y-1) - y^x < 0
using the lemma below. Hence... (consider x -> 1).
Lemma: If 0 < y < x < 1 then x^(1/(x-1)) < y^(1/(y-1)).
Todd Trimble
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