When is it valid to convert "$\le$" or "$\ge$" to "$=$"? I don't understand why the conversion of ≤ to = in this proof in Spivak's Calculus is legitimate. (The conversion in the previous line from = to ≤ makes sense to me.)  Could someone please explain or elaborate on how $(2)$ follows from $(1)$ below?

this proof is motivated by the observation that
$$
|a|=\sqrt{a^2} \text {. }
$$
We may now observe that
$$
\begin{align*}
(|a+b|)^2=(a+b)^2 & =a^2+2 a b+b^2 \\
& \leq a^2+2|a| \cdot|b|+b^2 \tag{1} \\
& =|a|^2+2|a| \cdot|b|+|b|^2 \tag{2} \\
& =(|a|+|b|)^2
\end{align*}
$$


 A: It is not converting anything. It is simply saying something akin to
$$1+1 \leq 2+1 = 3$$
and then concluding that $1+1 \leq 3$.
A: In case the existing answers are still not clear...

*

*The presentation $$
\begin{align*}
{}& (|a+b|)^2\\
={}& (a+b)^2 \\
={}& a^2+2 a b+b^2 \\
\leq {}& a^2+2|a| \cdot|b|+b^2 \tag{1} \\
={}& |a|^2+2|a| \cdot|b|+|b|^2 \tag{2} \\
={}& (|a|+|b|)^2
\end{align*}
$$ is asserting the conjunction of these 5 statements:
\begin{align*}
 (|a+b|)^2 &= (a+b)^2 \\
(a+b)^2 &= a^2+2 a b+b^2 \\
a^2+2 a b+b^2 &\leq a^2+2|a| \cdot|b|+b^2 \tag{1'}\\
 a^2+2|a| \cdot|b|+b^2 &= |a|^2+2|a| \cdot|b|+|b|^2 \tag{2'}\\
 |a|^2+2|a| \cdot|b|+|b|^2 &= (|a|+|b|)^2.
\end{align*}


*It is then natural to conclude (using transivity of = together with substitution of each side of the inequality) that  $$(|a+b|)^2 \le  (|a|+|b|)^2.\tag C$$
Note that $(C)$ is generally not explicitly asserted, but just tacitly understood by the reader.


*

*

*Lines $(1)$ and $(2)$ are not statements, and we are not asserting that $(1)$ implies $(2)$ (nor asserting that they are equivalent, for that matter):
\begin{align}
\leq {}& a^2+2|a| \cdot|b|+b^2 \tag{1} \\
={}& |a|^2+2|a| \cdot|b|+|b|^2. \tag{2} 
\end{align}

*Neither is fragment $(2)$ replacing fragment $(1);$ the ≤ symbol is not being dropped and subsequently ignored!



*On the other hand, given \begin{align}
a<b\\
b=c\\
c>d,
\end{align} it is invalid to draw any of the conclusions

*

*$a<d$

*$a=d$

*$a>d;$
in the absence of more data, any of these 3 statements might be true.
A: The reason the equality follows is that, for any $a,b \in \mathbb{R}$, we have
$$|a|^2 = a^2$$
Compare this against the usual definition of $|a|$:
$$|a| := \begin{cases}
a & a \ge 0 \\
-a & a < 0 \end{cases}$$
but if you square the result, in either case, you get $a^2$.
There is no "conversion from $\le$ to $=$" going on here; this is just another logical step in the proof. We know that $|a|^2=a^2$ and $|b|^2=b^2$, so the former line yields the latter.
