if $a,b$ such $a^2+b^2+4=2a+ab+2b$,find $a^2b$ let $a,b\in R$, and such
$$a^2+b^2+4=2a+ab+2b$$
Find $a^2b=?$
My idea: $$a^2-2a+1+b^2-2b+1+2+ab=0$$
$$(a-1)^2+(b-1)^2+ab+2=0$$
then I can't it
maybe this problem   can use inequality to solve it,Than you 

other solution: since
$$a^2-(2+b)a+b^2-2b+4=0$$
$$\Delta_{a}=(2+b)^2-4(b^2-2b+4)=-3b^2+12b-12=-3(b^2-4b+4)=-3(b-2)^2\le 0$$
so
$$b-2=0\Longrightarrow b=2\Longrightarrow a=2$$
