Fun Math Problem?

Question

If x and y are both positive real numbers, show that x^y + y^x > 1

I suppose I should clarify. Show for ANY positive real numbers x and y,

x^y + y^x > 1

By the way, this is not a homework problem. It's listed as a challenging problem on a mathmatics website. I thought people might get a kick out of it.

Answer 1

That's easy, if i could show you my hand i'd show you the one.

Answer 2

The problem is trivial for x>=1, y > 0 and y >= 1, x > 0
so I consider the square 0 < x < 1, 0 < y < 1

By symetry I need only consider 0 < y <= x < 1

If x = y, set g(x) = 2 x ^ x = 2 exp (x log x)

differentiating,
g'(x) = 2 > (1 + logx) exp (x log x) = 0 and x = 1/e

g(1/e) =(min on x=y0
=2 1/ (exp (1/e)) > 2/ e^1/2 [since e > 2] > 2/ (3 ^ 1/2)
[since e < 3] > 1. If min greater than 1, then g(x) > 1 for
0 < x < 1

Finally consider 0 < x < y <1, then x^y > x^x and y^x > x^x

x^y + y^x > 2 x^x > 1 by previous calculation,.

It may be possible to do 0 < x < y at one time.

Answer 3

The principle of complete induction states that if P(n) (for n=some real positive number) implies P(n+1) is true, then P(n) is true for any n>0. In other words,
let x=1 and y=1, then
for P(n+1)= (1+1)^1+(1+1)^1>1,
2+2>1
4>1, is true, so for P(1)=1^1+1^1>1, so 1+1>1,so 2>1, true
So x^y + y^x > 1 is true for any x,y=positive real number

Answer 4

We would have to take this problem in cases:
Case 1: x=1 and y=1
Then 1^1 + 1^1 = 1+1 = 2>1
Case 2: x=1 and y>1
Then 1^y + y^1 = 1+y >1
(Same for x>1 and y=1)
Case 3: x>1 and y>1
This case is trivially provable
Case 4: x=1 and y<1
Then 1^y + y^1 = 1+y >1
(Same for x<1 and y=1)
Case 5: x<1 and y<1
This is the hardest case and I can't think right away a direct proof, but it does work.

Answer 5

This problem can best be solved by 2 variable optimization..
we need optimise this function f(x,y)=x^y + y^x and prove that this can't be less than 1.
Now for this we need to solve simultaneously first derivative equations.
Here
df/dx = y*x^ (y-1) + y^x ln(y) [hope you can do it] this is partial derivative with respect to x only

also df/dy= x*y ^(x-1)+x^y * ln(x) , just change y with x here.
now we need to solve for df/dx=df/dy=0

take ln( y) = -y^(1-x)* x^(y-1) from first equation
ln( x) = -x^(1-y)* y^(x-1)
multiply these two..then ln(xy)= x^0*y^0=1
hence this is possible only when x=1 or 0 or vice versa for y.
then maxima or minima points are (1,0), (0,1).
whether they are maxima or minima can be found by the second derivative. See the details in this link.

But here it's minima.
So minimum value of this function f(x,y)=x^y + y^x = 1^0+0^1=1

Hence this function can not have a value less than 1 for any real x and y s which are >0.

Answer 6

lets do it in several cases


Case 1: x=1 and y=1
Then 1^1 + 1^1 = 1+1 = 2>1
Case 2: x=1 and y>1
Then 1^y + y^1 = 1+y >1
(Same for x>1 and y=1)
Case 3: x>1 and y>1
This case is trivially provable
Case 4: x=1 and y<1
Then 1^y + y^1 = 1+y >1
(Same for x<1 and y=1)
Case 5: x<1 and y<1
This is the hardest case and I can't think right away a direct proof, but it does work.

Answer 7

wait .. unless i'm mistaken .. couldn't the numbers be anything over 1? say x=1 and y=2 .. then 1^2=1 and 2^1 is 2 2+1= 3 and 3>1 .. or am i just confused?

Answer 8

Nikki,

You proved for all positive integers, not all positive real numbers. Thanks for playing.

This problem is solved by the fact that one of x^y or y^x must be ≥1 itself for any two positive real numbers.

Answer 9

sorry, but I am not doing your math homework. now get back to work and figure it out yourself.

Answer 10

thats easy
u go do it
well im jus answering this for two points