sorry the question is find all real $x,y$ such that $x(2y+1)\geq x^2 + 2y^2 + 1/2 $

solution the given statement is equivalent to $ (2y-x)^2 + (x-1)^2 <= 0 $ but by trivial inequality $a^2\geq 0 $ for all real $a$ hence $2y-x = x-1 = 0 $ and hence $ x=1$ and $ y= 1/2$

I know that it's been 8 years since this but could someone please explain why Quote: sorry the question is find all real $x,y$ such that $x(2y+1)\geq x^2 + 2y^2 + 1/2 $ is different from the original question?

Erratum wrote: I know that it's been 8 years since this but could someone please explain why Quote: sorry the question is find all real $x,y$ such that $x(2y+1)\geq x^2 + 2y^2 + 1/2 $ is different from the original question? The original problem is editted to the current one I think the poster recognized it because of #2