the minimum value of $a^2+b^2-a-\frac{2b}{3}$ using known standard inequalities If $a,b$ are real numbers satisfying $a+2b=3,$ then the minimum value of $a^2+b^2-a-\frac{2b}{3}$

Here $a+2b=3\implies a=3-2b$
$a^2+b^2-a-\frac{2b}{3}=(3-2b)^2+b^2-(3-2b)-\frac{2b}{3}$
$=9+4b^2-12b+b^2-3+2b-\frac{2b}{3}$
I diiferentiated it with respect to $b$ and put it equal to $0$ and then found $a$ by using the relation $a=3-2b$ to get the minimum value. 
But this is a long method.Can we find its minimum value by using known inequalities like AM-GM etc.

Hint : complete squares !can you see the answer?
