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

As a function of $a$, the minimum of the quadratic $4a^2+b^2+1-(2ab+2a+b)$ is $\frac34 (b - 1)^2 \ge 0$.
More precisely,
$$4a^2+b^2+1-(2ab+2a+b)=\left(2a-\frac{b+1}{2}\right)^2+\frac34 (b - 1)^2 \ge 0$$ 
