Why does = change to $\leq$ and then to = in this proof of |a+b| = |a|+|b|?

Question

From Spivak's Calculus.

This proof is motivated by the observation that |a| = $\sqrt {a^2}$. $\sqrt x$ denotes the positive square root of x; this symbol is defined only when x $\geq 0$. We may now observe that.

$$ (|a + b|)^2 = (a+b)^2 = a^2 + 2ab + b^2 \leq a^2 + 2|a| |b| + b^2 \\ =|a|^2 + 2|a| |b| + |b|^2 =(|a| + |b|)^2 $$

I don't understand why the $\leq$ appears in line 2, and then changes to $=$ in line 3. Thank you.

Answer 1

It doesn't really "change" in the sense that the equality is replaced by an inequality or vice versa; what you have here is a chain of equations and an inequality, which say that $(|a+b|)^2 = (a+b)^2$ and $(a+b)^2 = a^2+2ab+b^2$ and $a^2+2ab+b^2\leq a^2+2|a|*|b|+b^2$, etc.

Answer 2

In the second line, we have $ab \leq |a||b|$. This is true because if $ab\leq0$ then $ab\leq 0 \leq |a|b|$. If $ab\geq 0 $ then $ab=|ab|=|a||b|$. In any case it holds.

In the third line we use that $a^2=|a^2|$ and $b^2=|b^2|$ which is true because $a^2,b^2 \geq 0$. Notice we used that on the very first step as well, but the other way around.

Now, the inequality doesn't vanish. The end result is $(|a+b|)^2 \leq (|a|+|b|)^2$.

What we have done is essentially this: $A \leq B = C \Rightarrow A \leq C$.

Answer 3

In the line 2 you have dropped a "2" since it is $|a|^2+2|a||b|+|b|^2$. Also, in the line 2, $\le$ appears because $ab\le |ab|$. For the line 3 it is "=" since $a^2=|a|^2$ and $b^2=|b|^2$.