Question

Proof that x^y + y^x > 1 for all x,y > 0

Answer 1

x,y>0
so , x^y>0 and y^x>0
both of them cannot be 0
and so it is x^y + y^x > 1
hence proved

Answer 2

@Dhashi - how do you conclude x^y+y^x>1 if each component is greater than 0?

Answer 3

that's not a proof.

Answer 4

why can't they both be 1/4?

Answer 5

here is a nice one using bernoulli
http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1213001311

Answer 6

It was a nice try though ;)

Answer 7

positive quantity raised to positive quantity always positive

Answer 8

@jamesj, for some reason i feel like lighting up a marlboro

Answer 9

AravindG and DHASHNI propably didnt see that the RHS was 1, not 0

Answer 10

It looks like a very interesting problem - I wonder if there is a simple way of proving it (apart from the one @satellite73 showed in the link)?

Answer 11

just wondering if we could use logs some how?

Answer 12

@sat73, Phillip Morris paid me to use that photo.

I can't seem to be able to read the proof you linked to, and it's being difficult about registering. Would you mind taking a quick screen shot of the text and posting it?

Answer 13

James use your mouse.

Answer 14

Hahaha asnaseer quite a good result using logs hehehe if x and y are 1 tehn using log we will remain with 0>0 and the question itself may be wrong it might be > or equal to 1

Answer 15

ah, select the text, got it.

Answer 16

mouse over, or you can look here i think it is ok
http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=print;num=1213001311

Answer 17

hmmmm sry sry

Answer 18

This is the text from that site:
Assume that both x and y lie in (0,1) (otherwise, we are done).
Now wlog assume that y1/(1-y) >= x1/(1-x)
Note that, this implies that
xy-1/yx-1 >= 1 -- ( A )
Now Consider S = (xy + yx)1/x
= [yx(1 + xy/yx)]1/x
= y(1 + xy/yx)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 + xy-1/yx)
= y + xy-1/yx-1 >= y + 1
(by A)

Thus S >= 1 + y > 1
Since S > 1, Sx > 1 and we are done.

Answer 19

This is also from the same site:
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

Answer 20

Got one thing wrong with logs i have assumed wrong sry guys there's no prob with the question

Answer 21

Y cant we guys just differentiate and find the minimum value of the function???????

Answer 22

Using some Tyler expansion or that kinda thing for the function of two variables

Answer 23

Asnaseer I need ur comment here

Answer 24

I tried differentating, but its rather messy, so I didnt want to go on that. The question is something my friend asked me on facebook couple of weeks ago, and I couldnt solve.

Answer 25

pls help me in physics after this guys.come to physics section????

Answer 26

@Atchyut - what do you want me to comment on?

Answer 27

@asnaseer: I didn't understand:

Now wlog assume that y1/(1-y) >= x1/(1-x) what's is wlog?

Answer 28

hmmm - I just copied the text from the site that @satellite73 gave above.

Answer 29

I know, but you can you use your amazing mind to figure it out though ;)

Answer 30

wlog means without loss of generality

Answer 31

thx @JamesJ

Answer 32

argghhh that's silly abbreviation, I was thinking some sort of mathematical function or something.