Is there an accepted strategy when tackling proofs involving inequalities? 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!
 A: In general, there is no method that will solve any inequality. Inequalities need to be tackled case by case, although we can simplify particular forms given prior experience and knowledge. General mathematical intuition and perserverance will go a long way toward unfamiliar forms.
A specific strategy is to change your variables so the condition is something nice. This usually makes the problem much easier to read and digest. Here it doesn't change too much but it's a good general strategy to keep in mind, so I will demonstrate it anyways.
Let $b = c-1$. This means if $b \ge -1$, then $c \ge 0$. Now we can rewrite the inequality as:
$$
\frac{4(c-1)^2+c}{4|c-1|} \ge \sqrt{c}
$$
Next, notice that there is a $|c-1|$ in the denominator and a $(c-1)^2$ in the numerator. This is very good because it will be very easy to reduce.
$$
\frac{4(c-1)^2}{4|c-1|} + \frac{c}{4|c-1|} \ge \sqrt{c} \\
|c-1| + \frac{c}{4|c-1|} \ge \sqrt{c}
$$
Now multiply out the denominator:
$$
4(c-1)^2 + c \ge 4|c-1|\sqrt{c} \\
4(c-1)^2 - 4|c-1|\sqrt{c} + c \ge 0
$$
Setting $0$ on one side is very useful here because we know that $a^2 \ge 0$ for all real $a$. We can try to see if the left side is a perfect square:
$$\left(2|c-1| - \sqrt{c} \right)^2 \ge 0$$
It is in fact a perfect square! This means that as long as the expression in parentheses is real, we have a solution. The only way in which this works is if $\sqrt{c}$ is positive, or in other words, $c \ge 0$. We remove the $c = 1$ singularity and we are done!
EDIT: @pidgeon's comment is excellent, because it takes an inequality that is true for an entire domain and applies it directly. You can take that inequality and immediately apply it to a problem if you can find the clever substitutions. This will be much faster than my more brute force approach, but it is the method I use when I don't have a better one.
Another really great inequality to always keep on hand is the AM-GM-HM inequality, which simply states that for any set of positive numbers, the arithmetic mean is at least the geometric mean, which is at least the harmonic mean.
A: 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.
A: By AM-GM
$$\frac{4b^2+b+1}{4|b|}=|b|+\frac{b+1}{4|b|}\geq2\sqrt{|b|\cdot\frac{b+1}{4|b|}}=\sqrt{b+1}.$$
