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

Let $x=(2a,b,1)$ and $y=(1,2a,b)$. Then by, Cauchy-Schwarz,
$2ab+2a+b= <x,y> \le ||x||*||y||=\sqrt{4a^2+b^2+1}*\sqrt{4a^2+b^2+1}=4a^2+b^2+1$
