Prove $4a^2+b^2+1\ge2ab+2a+b$ Prove $4a^2+b^2+1\ge2ab+2a+b$
$4a^2+b^2+1-2ab-2a-b\ge0$
$(2)^2(a)^2+(b)^2+1-2ab-2a-b\ge0$
Any help from here? I am not seeing how this can be factored

High school solution: 
$$
4a^2+b^2+1-2ab-2a-b=\frac{(2a-1)^2+(2a-b)^2+(b-1)^2}{2}\ge 0.
$$
