How should I think when combining multiple inequalities? When reading/writing papers, I have always find it not obvious when two or more inequalities are combined. For example, taken from my current research
$$\text{Pr}(X \le ab) \le -a (1-p)^{-N} (1 - (1-p)^{-N/2})$$
(here we have $N > 0$, an integer)
Here, I want to find $p$ that will eliminate all the $N$ factors, e.g., $p = 1/N$. But I now will need to plug either one of these inequalities
$$(1 - p) \le e^{-p}$$
$$(1 - p) \ge e^{-p + p^2}$$
I need a rather long time to convince myself to use the second inequality for $(1-p)^{-N}$ and the first for $(1-p)^{-N/2}$.
Is there a technique, shorthand, or something that helps this kind of thinking? Alternatively I would appreciate it if you would share how you approach this kind of problems, so that I may learn how too.
 A: To work with a slight simplification of your example, let's say you want to find an upper bound for
$$(1-p)^{-N}((1-p)^{-N/2}-1)$$
The general idea is you want to break things down into steps as much as possible. Introducing some algebra will also be helpful.  Here we have a product of 2 positive terms, so it will suffice to get an upper bound for each of these.  (If you want an upper bound for a negative number, it's maybe mentally easiest to think about looking for a lower bound of its negative.)  I rewrote your example on purpose to look like a product of two positive terms, so you don't need to worry about flipping inequalities with multiplication by negative numbers.
First consider $(1-p)^{-N}$, which I think is helpful to write as $\frac 1{(1-p)^N}$.  Now let's introduce some algebra.  You want a bound $X (> 0)$ such that
$$ \frac 1{(1-p)^N} \le X \iff \frac 1X \le (1-p)^N \iff \sqrt[N]{\frac 1X} \le (1-p).$$
Thus you want to use a lower bound for $(1-p)$ here.  The most straightforward thing to do is set $\sqrt[N]{1/X}$ equal to your lower bound and solve for $X$.
Similarly, next you want an upper bound $Y$ for $((1-p)^{-N/2}-1)$.  This means you want $Y$ so that
$$ (1-p)^{-N/2} \le Y + 1.$$
This is similar to the above case, and an upper bound for the product is $XY$.
