I'm sure there's an easier way that's not so messy, but since this is in calculus, I'm using a calculus method. And the actual math can get really ugly, so I'm just gonna leave out a bunch of ugly math.

Consider z=x^y+y^x-1. To find the extrema, we must find dz/dy=dz/dx=0 (partial derivates).

dz/dx=y*x^(y-1)+lny*y^x=0
dz/dy=x*y^(x-1)+lnx*x^y=0

By symmetry, you can see that there should be a solution with x=y. Plugging that in gets a solution x=y=1/e. To find out if there are other solutions, rewrite this as

y*x^(y-1)=-lny*y^x
x*y^(x-1)=-lnx*x^y

Multiplying and dividing by x^y*y^x gets lnx*lny=1. Therefore, x=exp(1/ln(y)). Substituting this back into the the equations for dz/dx will yield

y*exp((y-1)/lny)+lny*y^(exp(1/lny))=0.

This equation only has one solution, y=1/e. You can see this from a plot or a more formal analysis by breaking into regions where it increases and regions where it decreases.

The solution x=y=1/e can turn out to be a maximum or a minimum or a saddle point. This can be verified by the Second Derivative Test. It turns out to be a minimum.

Plugging in x=y=1/e yields z=2*exp(-1/e)-1=0.384401. x^y+y^x>=1.384401. Hence, x^y+y^x>1.