How does this proof of Theorem 1 in Spivak's Calculus work?

Question

Hello I just started Spivak's Calculus and I've come across something I don't really understand. What is trying to be proved is $$ |a+b|\leq|a|+|b|\,. $$ The proof is based on the observation that $|a|=\sqrt{a^2}$. So: $$ \begin{align*} (|a+b|)^2 = (a+b)^2 &= a^2+2ab+b^2 \\ &\leq a^2+2|a|\times|b|+b^2 \\ &= |a^2|+2|a|\times|b|+|b^2| \\ &= (|a|+|b|)^2 \\ \end{align*} $$ Now what I don't understand is the change in relationship from line 1 to 2 ($= to \leq$) and from line 2 to 3 ($\leq to =$).

I attach the page because it's clearer there (I highlighted the proof in yellow). Actual proof from the book

Answer 1

The lines are read left to right and top to bottom. So the symbol ≤ on line 2 relates the last expression of line 1 to the next expression on line 2. Similarly, the symbol = on line 3 relates the last expression of line 2 to the next expression on line 3. The same thing could be written in one line as follows: \begin{equation} \begin{split} (|a+b|^2) = (a+b)^2 &= a^2+2ab+b^2 \leq a^2+2|a|\times|b|+b^2 = |a^2|+2|a|\times|b|+|b^2| = (|a|+|b|)^2, \end{split} \end{equation} but it is easier on the eyes to write each new expression on a new line.

Answer 2

For any number $x \in \mathbb{R}: x \leq |x|$

Hence, $ab \leq |ab| = |a||b| \implies 2ab \leq 2|a||b| \implies a^2 + 2ab + b^2 \leq a^2 + 2 |a||b| + b ^2$

Answer 3

$$a^2+2ab+b^2 \leq a^2+2|a|\times|b|+b^2 $$ because $$ab \le |a||b|$$

Also $$a^2+2|a|\times|b|+b^2 = |a^2|+2|a|\times|b|+|b^2| $$

because $a^2 = |a|^2 $ and $b^2 = |b|^2 $