Maximum value of $x$ when equality is given $$ x + y = \sqrt{x} + \sqrt{y} $$ 
Find maximum value of $x$. $x$ and $y$ are reals.

Try this method of completing the square.
Let $a=\sqrt x$ and $b=\sqrt y$ so that $a^2+b^2=a+b$ and $4a^2-4a+1+4b^2-4b+1=2$ ie $(2a-1)^2+(2b-1)^2=2$
