>>200
Let u=x-y, v=x+y=1, then x=(1+u)/2,y=(1-u)/2 . -1≦u≦1
We get f(u)=((u+1)/2)^(1-u)+((1-u)/2)^(1+u).
=g(u)+k(u)
g(u)=(u+1)/2)^(1-u), k(u) = ((1-u)/2)^(1+u)

We find g(u) is a monotone decreasing function and k(u) a monotone increasing function.
( g'(u) <0, k'(u)>0)
and f(u) is a symmetric function.

f(1)=f(-1)=f(0)=0
f'(0)=0
f(-1/2)=f(1/2)=1/8+3^(1/2)/2<1, (0.991025)
These relation show that f(u) decreases from u=-1 and increase somewhere (-1,0).
The behavior of f(u) over (0,1) is symmetric.
Thefore we can say f(0) is a peak candidate with terminal f(-1) and f(1).
SO we can say f(u)<=1.
Q.E.D