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        riddles >> medium >> Prove x^y+y^x>1
        
(Message started by: cool_joh on Jun 9^th, 2008, 1:48am)

Title: Prove x^y+y^x>1
Post by cool_joh on Jun 9^th, 2008, 1:48am

Prove that the inequality x^y+y^x>1 holds for any positive reals x and y.

Title: Re: Prove x^y+y^x>1
Post by Aryabhatta on Jun 9^th, 2008, 6:18pm

This below seems to work, but I have a strange feeling something is wrong somewhere...

[hide]

Assume that both x and y lie in (0,1) (otherwise, we are done).

Now wlog assume that y^1/(1-y) >= x^1/(1-x)

Note that, this implies that

x^y-1/y^x-1 >= 1 -- ( A )

Now Consider S = (x^y + y^x)^1/x

= [y^x(1 + x^y/y^x)]^1/x

= y(1 + x^y/y^x)^1/x

Now since 1/x > 1, by Bernoulli's inequality that (1+x)^r >= 1 + rx for x > -1 and r >= 1

we have that

S >= y(1 + x^y-1/y^x)
= y + x^y-1/y^x-1 >= y + 1
(by A)

Thus S >= 1 + y > 1

Since S > 1, S^x > 1 and we are done.

[/hide]

Title: Re: Prove x^y+y^x>1
Post by wonderful on Jun 12^th, 2008, 10:31pm

Here is another solution:

[hide]Case 1: Either x,y >= 1 then x^y >=1 or y^x >=1

=> x^y + y^x >1

Case 2: x,y<1

By Bernoulli inequality, we have:
x^y = [1 + (x-1)]^y >= 1 + y(x-1)
=> x^y + y -1 >= xy

Likewise:
y^x + x-1 >= xy

Thus, if we can show 2xy+1>= x+y the x^y + y^x > 1.

This is true since (x-1)(y-1)+xy>0 given <0x,y<1[/hide]

Have A Great Day!

Title: Re: Prove x^y+y^x>1
Post by Aryabhatta on Jun 13^th, 2008, 1:12am

on 06/12/08 at 22:31:48, wonderful wrote:

Here is another solution:

Case 2: x,y<1

By Bernoulli inequality, we have:
x^y = [1 + (x-1)]^y >= 1 + y(x-1)
=> x^y + y -1 >= xy

wonderful, in case the exponent is between 0 and 1, the Bernoulli inequality is reversed, i.e

(1+x)^r <= 1 + rx, so the application here is incorrect.

Title: Re: Prove x^y+y^x>1
Post by wonderful on Jun 13^th, 2008, 1:37pm

Thanks Aryabhatta! You are right.

Have A Great Day!