# Spivak Calculus 3rd ed. $|a + b| \leq |a| + |b|$

I'm working through the first chapter of Michael Spivak's Calculus 3rd ed.

Towards the end of the chapter he proves $|a + b| ≤ |a| + |b|$ using the observation that $|a|= \sqrt{ a^2 }$ when $a$ is $≥ 0$ .

$|a + b| ≤ |a| + |b|$

$$(|a + b|)^2 = (a + b)^2$$ $$= a^2 + 2ab + b^2$$ $$≤ a^2 + 2|a| |b| + b^2$$ $$= |a|^2 + 2|a| |b| + |b|^2$$ $$= (|a| + |b|)^2$$

I am unsure about what's going on with the equality sign. How does it go from $=$ to $≤$ on line 3 when a and b are changed to their absolute value and back to $=$ again on line 4 when $a^2$ and $b^2$ are changed to their absolute values?

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When there is a chain of expressions like this, an = sign only means equal to the immediately preceding line, not to all the previous lines. –  Ted Aug 17 '12 at 6:20
Looking at this,I feel you are unable to see why $2|a|*|b| \ge 2ab$.To see why that is being said, just take a=-3 and b=3.That's why the modulus sign is there.Michael Joyce has already posted an answer and as he pointed out , the key is $x\leq |x|$. –  user37450 Aug 17 '12 at 6:33

To extend Ted's comment, Spivak is claiming that: $$a^2 + 2ab + b^2 \leq a^2 + 2|a| |b| + b^2$$ (this follows because $x \leq |x|$ for every $x \in \mathbb{R}$) and that: $$a^2 + 2 |a| |b| + b^2 = |a|^2 + 2 |a| |b| + |b|^2$$ (this follows because $x^2 = |x^2| = |x|^2$ for all $x \in \mathbb{R}$ since $x^2 \geq 0$ for all $x \in \mathbb{R}$).

You can then "string these statements together" to conclude that $$a^2 + 2ab + b^2 \leq |a|^2 + 2 |a| |b| + |b|^2.$$ (The general principle being $x \leq y$ and $y = z$ implies that $x \leq z$.)

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Thanks Michael, that connection makes this proof easier to understand –  synk113 Aug 17 '12 at 6:33

Let $a=r_1(\cos A+i\sin A)$ and $b=r_2(\cos B+i\sin B)$

So, $|a|=r_1$ and $|b|=r_2$

Now, $|a+b|=\sqrt{(r_1\cos A + r_2\cos B)^2+(r_1\sin A + r_2\sin B)^2}$

$=\sqrt{r_1^2+r_2^2+2r_1r_2\cos(A-B)}$

$≤\sqrt{r_1^2+r_2^2+2r_1r_2}\$ as $\cos(A-B)≤1$ for real A,B

$=r_1+r_2=|a|+|b|$

Also observe, $|a+b|=\sqrt{r_1^2+r_2^2+2r_1r_2\cos(A-B)}≥\sqrt{r_1^2+r_2^2-2r_1r_2}=||r_2|-|r_1||$ as $\cos(A-B)≥-1$ for real A,B

$|a+b|≥||b|-|a||$

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Perhaps taking $x=a+ib$ and $y=c+id$ would have done the job as well. –  user37450 Aug 17 '12 at 6:49