How could I know where to begin a proof a priori? I am reading "Mathematical Thinking: Problem-Solving and Proofs" by D'Angelo and West.
On chapter 1, inside the "Elementary Inequalities" section, we have proposition 1.3 "Triangle Inequality".
I can follow perfectly and I  understand the justification behind each step. But there is something that is irking/puzzling me...The proof begins with:
"We start off with the inequality $2xy \le 2 |x| |y| $ "
If I had attempted beforehand to prove the proposition on my own, how could I have known this and that that inequality had some relationship with this proposition?
Where does this come from? (There is no previous text showing this inequality)
I assume the author is not...well, the author of this proof. Yet whoever came up with this proof in the first place had to see some sort of connection. Otherwise, we would be beginning proofs with a random initial step, right?
 A: Usually, what one writes as a formal proof is not the same way the person thought of it. Because usually, the way we "think" of the problem is usually in the opposite direction of the logic of the proof. Let me try to illustrate with an example.
Suppose we want to prove that for every $a,b \geq 0$, we have $\sqrt{ab} \leq \dfrac{a + b}{2}$ (the AGM inequality). I remember exactly how I proved this statement the very first time I saw it. What I did was something very bad logically: I started by assuming the conclusion, but from an intuitive standpoint, if you have absolutely no clue whatsoever, then this approach could be somewhat helpful, because it atleast gives you something to work with.
So, I started by squaring both sides, $ab \leq \left(\frac{a+b}{2}\right)^2 = \frac{a^2+2ab + b^2}{4}$. Then, multiply by $4$ to get $4ab \leq a^2 + 2ab + b^2$. Or rearranging, $0\leq a^2-2ab + b^2 = (a-b)^2$, and this last statement is certainly true. Of course, logically whatever I have done amounts to nonsense because I haven't done/proven anything directly addressing the proposition.
But, by doing this preliminary work, you now atleast have an idea of what to try, and where to start from; the goal is to try to "undo" each step (in this very simple case, it turns out to be possible). So, if I were to write a formal proof, it would be something like:

Suppose $a,b \geq 0$. We start with the inequality $0 \leq (a-b)^2$, which is always true. Next, we expand it out to get $0\leq a^2-2ab+b^2$. Now, we add $4ab$ to both sides, to get $4ab \leq a^2 + 2ab + b^2$. Now, divide by $4$, and factor the numerator of the RHS to get $ab \leq \left(\frac{a+b}{2}\right)^2$.
Now, in this final stage, because $a\geq 0$ and $b\geq 0$ by hypothesis, the product $ab \geq 0$ as well. Hence, we can "take square roots" on both sides to get $\sqrt{ab} \leq \frac{a+b}{2}$.

And of course, if this is the very first time seeing it, then you might be surprised/questioning how did I even think to start with $0\leq (a-b)^2$. Well, the answer is I didn't get it out of thin air; it was an educated guess based on my "scratch work". But as you learn and read more, you usually pick up more and more tricks/ideas/techniques, so that if you encounter a completely new problem, you'll have a larger number of "educated guesses" to make. If one doesn't pan out, try another one. It's really a matter of practice.

As for your actual inequality $2xy \leq 2 |x||y|$, I'm not sure the context of the problem, but it's very easy to see that for any $a\in \Bbb{R}$, we have $a\leq |a|$ (it's a simple two-case verification). Now, just plug in $a=2xy$. Then, we have $2xy \leq |2xy|=2 |x||y|$.
