<s>[INST] 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 
 [/INST] The expression can be written as,
$a^2-ab+b^2-2a-2b+4=\frac{3}{4}(a-b)^2 + \frac{1}{4}(a+b-4)^2=0$
Since, $(a-b)^2\ge0$ and $(a+b-4)^2\ge 0$,
So, the only possible real solutions are $a-b=0$ and $a+b-4=0$
That is $a=b=2$.
$a^2b=8$.
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