11I've kinda solved this, but the solution is not quite elegant....So I'm prefering to wait a bit more, till others kill this with their fine-touches...my soln is be there at the last ...
For positive x and y, establish the following inequality
x^y + y^x>1.
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13 Answers
1111Well, I found out another short solution....
Snce the only case to be dealt with is when both x,y are <1, we have x=cosa and y=sinb where b>a....
cossinba>cos2a
sinb>sina
Thus sincosab>sin2a....
Adding the 2 we get
cossinba+sincosab>1.....Putting values, we have
xy+yx>1....
66Soumik, b>a does not necessarily means sin b> sin a. So u have to consider different intervals of a and b.
11What if i take both a and b in the 1st quadrant.....
For all x, satisfying 0<x<1, we can always find such a and b.....(can't we?)
Actually I take the minimum value of 'b' satisfying cosb=x.....and same for a....coz that still proves for all x and y bet 0 & 1......Am i correct or wrong?
Pls answer...
39i have a solution for 0<x,y<1 but not quite sure that can be generalised.
kaymant sir /prophet sir if u throw some light on this, i can give my proof..
11
Ah, even if u raise that argument, that's done.....
The diagrm illustrates the case....
For the worst case, let a>k, so that sinb<sina....
b=(90+n), so x=cosa=cos (90-m)=sinm....
y=sinb=sin(90+n)=cosn....since a>k, n>m...
Now proceeding as before...
sincosmn>sin2n
cosm>cosn, as both are acute angles...
cossinnm>cos2n
Adding the 2 we get the result....
(This is the only case where b>a yet sinb<sina, where both sina and sinb>0)....
1here is my solution http://www.4shared.com/file/134549494/b0bd38d6/RAHUL_DULGGAL.html
1a little change
http://www.4shared.com/file/137739217/2be0f850/RAHUL_DULGGAL.html
62I dont know.. but why do i have a feeling that I have heard your name somewhere?
1xy + yx achieves minimum when both are equal.......as in case a.m>=g.m equality is achieved wen all terms are equal
2xy/2 =minimum=m........
m=xy
minimum value of this >1....as y >0
66Is it? With the values you gave, both xy and yx are of the form 00 and the limiting value is 1 for each. So the sum is actually close to 2 which is more than 1. To be more precise, here is an output with Mathematica.