Math proof... any idea?

First I think you've left out an assumption:a and b are not both equal to each otherIf a and b = 1 then ((a+b)^2)/(4*a*b) = 1 (not > )(a+b)^2 = a^2 + 2ab + b^2
4ab = 2(2ab)
subtract 2ab from both numerator and denominator
a^2 + b^2 ~ 2ab
If a = b then a^2 = b^2
so the numerator becomes
2a^2 and the denominator becomes 2aa = 2a^2 so the quotient is 1.

If a and b are not equal then one is greater than the other by a positive amount say d.This is symmetrical so let's make a = b + dnow we have
(b + d)^2 + b^2 > 2(b+d)b
b^2 +2bd + d^2 + b^2 > 2b^2 + 2bd
and d^2 is > 0 as defined
so the left is > the right
It is true when not equal a and b