TITLE: Is there an accepted strategy when tackling proofs involving inequalities?
QUESTION [5 upvotes]: I am new to mathematical proofs, and I am not quite sure how to tackle problems of that nature. 

Here is the problem:

Prove that if $b≥-1$, $b≠0$, then $\frac{4b^2+b+1}{4|b|} ≥ \sqrt{b+1}$ 
I tried to rearrange the inequality on the right like so,
 $\frac{4b^2+b+1-4|b|\sqrt{b+1}}{4|b|} ≥ 0$

but this didn't really help me much. I am aware that I need to use the fact
that $b≥-1$ to proceed further, however, I don't know when it should come into play.

I am also not exactly sure if I always need to start working with the latter part ($\frac{4b^2+b+1}{4|b|} ≥ \sqrt{b+1}$) when dealing with proofs of this kind?
Thanks a lot in advance!

REPLY [3 votes]: Also, since both sides are nonegative, you can square it and get (recall $|x|^2 =x^2$)
$$(4b^2+b+1)^2\geq 16b^2(b+1)$$
which is the same as $$ 16b^4+b^2+1+8b^3+8b^2+2b\geq 16b^3+16b^2$$
or  $$ 16b^4+b^2+1-8b^3-8b^2+2b\geq 0$$
or $$(-4b^2+b+1)^2\geq 0$$ which is true.