TITLE: Prove $4a^2+b^2+1\ge2ab+2a+b$
QUESTION [5 upvotes]: 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

REPLY [3 votes]: Multiply by $2$, then we get:
$$8a^2+2b^2+2 \ge 4ab+4a+2b$$
and rearranging we get:
$$(4a^2-4ab+b^2)+(b^2-2b+1)+(4a^2-4a+1) \ge 0$$
and then
$$(2a-b)^2+(b-1)^2+(2a-1)^2 \ge 0$$

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