without loss of generality we can suppose \[y\ge x \]because the expression is symmetric
assume that \[y=x+a\]with \(a\ge 0\)
Okay! I'll try from here!!
wait ... im afraid i made a mistake :/
Don't be afraid :P \[4xy-x-y=4x(x+a)-x-(x+a)=4x^2+(4a-2)x-a\]
\[\Delta=(4a-2)^2 - 4(4)(-a)=0\]\[16a^2 - 16a + 4 + 16a=0\]\[16a^2 + 4 =0\]Well well well...
4xy-x-y = (x+y)^2 - (x-y)^2 - (x+y) = (x+y)(x+y-1) - (x-y)^2 i am not too sure if that'd help! :P
I've never heard of it! :(
Forgive me for closing this question, even though it's not yet solved!