Prove $\frac{4b^2+b+1}{4\left|b\right|}\ge \sqrt{b+1}$, $b\ge -1,\:b\ne 0$

Question

I've literally tried everything for this question. I think I'm supposed to use the Arithmetic-Geometric Mean Inequality (AGM) but I don't know how to apply AGM to this question so I tried algebraic manipulation and it feels like I'm getting nowhere.

$\frac{4b^2+b+1}{4\left|b\right|}\ge \sqrt{b+1}$

$\frac{\left(4b^2+b+1\right)^2}{16b^2}\ge b+1$

$b^2+\frac{b}{2}+\frac{9}{16}+\frac{1}{8b}+\frac{1}{16b^2}\ge b+1$

$b^2-\frac{b}{2}-\frac{7}{16}+\frac{1}{8b}+\frac{1}{16b^2}\ge 0$

$\frac{16b^4+2b-7b^2-8b^3+1}{16b^2}\ge 0$

$16b^4+2b-7b^2-8b^3+1\ge 0$

I don't know where to go from here.

I don't want a complete answer, just give me a hint, thanks.

EDIT: I think I figured it out:

AGM = $\frac{x+y}{\:2}\ge \sqrt{xy}$ = $\frac{x+y}{\:2}\ge \sqrt{x}\sqrt{y}$

Let x = $4b^2$ and y = $b+1$

Then we have, by AGM:

$\frac{\left(4b^2\right)+\left(b+1\right)}{2}\ge \sqrt{4b^2}\sqrt{b+1}$

$\frac{4b^2+b+1}{2}\ge 2\left|b\right|\sqrt{b+1}$

$\frac{\frac{4b^2+b+1}{2}}{2\left|b\right|}\ge \sqrt{b+1}$

$\frac{4b^2+b+1}{4\left|b\right|}\ge \sqrt{b+1}$, as needed. QED

Answer 1

Hint: Remember that $x^2 =|x|^2$ and write LHS as $${4b^2\over 4|b|} +{b+1\over 4|b|}$$