When is it valid to convert "$\le$" or "$\ge$" to "$=$"?

Question

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 $$ \begin{align*} (|a+b|)^2=(a+b)^2 & =a^2+2 a b+b^2 \\ & \leq a^2+2|a| \cdot|b|+b^2 \tag{1} \\ & =|a|^2+2|a| \cdot|b|+|b|^2 \tag{2} \\ & =(|a|+|b|)^2 \end{align*} $$

Answer 1

It is not converting anything. It is simply saying something akin to

$$1+1 \leq 2+1 = 3$$

and then concluding that $1+1 \leq 3$.

Answer 2

In case the existing answers are still not clear...

1. 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*}

2. 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.

3. • 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!

4. 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.

Answer 3

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.