Show that if $a$ and $b$ have the same sign then $|a + b| = |a| + |b| $, and if $a$ and $b$ have opposite signs then $|a+b| < |a| + |b|$

Question

I'm considering the different cases for $a$ and $b$

Case 1) $ a\geq 0$ and $ b\geq 0$

Given both terms are positive, $ a + b \geq 0 $ $$ |a+b| = a + b = |a| + |b|\\ $$

Case 2) $ a< 0$ and $ b< 0$

Given both terms are negative, $ a + b < 0 $ $$ |a+b| = -(a + b) = -a -b = |a| + |b|\\ $$

Case 3) $ a> 0$ and $ b< 0$

Case 3a) $ a + b \geq 0 $

Knowing that $ b < -b $ because $ b < 0 < -b $: $$ |a+b| = a + b < a - b = |a| + |b| $$ Case 3b) $ a + b \leq 0 $

Knowing that $ -a < a $ because $ -a < 0 < a $: $$ |a+b| = -(a + b) = -a - b < a - b = |a| + |b| $$

Case 4) $ a< 0$ and $ b> 0$

Case 4a) $ a + b \geq 0 $

Knowing that $ a < -a $ because $ a < 0 < -a $: $$ |a+b| = a + b < -a + b = |a| + |b| $$

Case 4b) $ a + b \leq 0 $

Knowing that $ -b < b $ because $ -b < 0 < b $: $$ |a+b| = -(a + b) = -a -b < -a + b = |a| + |b| $$

Is my proof correct?

I feel like it's somewhat incomplete as I'm not considering when either $a$ or $b$ are equal to zero in the cases 3 and 4.

Answer 1

Dealing with absolute values, the general strategy is: if you can avoid the cases method, do it.

Here , we have a typical example of this strategy: as both expressions are non-negative numbers, we compare them by comparing their squares, using that $|x|^2=x^2\,$: $$|a+b|^2=a^2+2ab+b^2,\enspace \text{ and }\enspace (|a|+|b|)^2=a^2+2|a||b|+b^2,$$ so that comparing $|a+b|$ and $|a|+|b|$ amounts to comparing $ab$ and $|a||b|=|ab|$, namely $$|a+b| <|a|+|b|\iff ab<|ab| \iff ab<0,$$ i.e., in ordinary language $a$ and $b$ have different signs.

Answer 2

I'll suggest some techniques for being briefer, which helps you know if you've covered all bases.

Since $|a+b|-|a|-|b|$ is unchanged when multiplying $a,\,b$ by $-1$, for the first problem we can assume without loss of generality that their common sign means they're $\ge0$, reducing the equation to the trivial $a+b=a+b$. The same invariance-under-changing-sign logic tells us the second problem only requires us to consider the case $a\ge 0>b$, whence$$(|a|+|b|)^2-|a+b|^2=a^2+b^2+2a|b|-a^2-b^2-2ab=4a|b|>0.$$

Answer 3

Something similar to yours:

The first part was, indeed, easy.

For the second part:

Without loss of generality, take $|a|>|b|~~~~~~~~~(*)$

• $a>0, b<0 \overset{(*)}\implies a+b>0$. Then, indeed, $\underbrace{|a+b|}_{=a+b} < \underbrace{|a|}_{=a}+\underbrace{|b|}_{=-b}$ since $b<-b$.
• $a<0, b>0 \overset{(*)}\implies a+b<0$. Then, indeed, $\underbrace{|a+b|}_{=-a-b} < \underbrace{|a|}_{=-a}+\underbrace{|b|}_{=b}$ since $-b<b$.

Answer 4

Similar to J.G.'s and Bernard's answers, we have

$$(|a|+|b|)^2-|a+b|^2=(a^2+2|ab|+b^2)-(a^2+2ab+b^2)=2(|ab|-ab)= \begin{cases} 0&\text{if }ab\ge0\\ 4|ab|&\text{if } ab\lt0 \end{cases}$$

Thus $(|a|+|b|)^2=|a+b|^2$ if $a$ and $b$ have the same sign, and $(|a|+|b|)^2\gt|a+b|^2$ if $a$ and $b$ have opposite signs. Since $|a|+|b|$ and $|a+b|$ are both nonnegative, we can take the square root on both sides of things, concluding that $|a|+|b|=|a+b|$ if $a$ and $b$ have the same sign and $|a|+|b|\gt|a+b|$ if $a$ and $b$ have opposite signs.

Note, this proof uses the absolute-value identities $|x|^2=x^2$ (for real numbers $x$) and $|x||y|=|xy|$.

Added later: In answer to the OP's question, "Is my proof correct?", it's true that some cases with $ab=0$ have been ignored, but I would say that the statement of the problem itself assumes $ab\not=0$, since it talks about $a$ and $b$ having the same or opposite sign, suggesting neither one is equal to $0$, which you might say lacks a sign. (There is, of course, a sign function, which is $+1$ for positive values, $-1$ for negative values, and $0$ for $0$. But if you interpret the problem as conditioning things on whether sgn($a)=\,$sgn($b$) or sgn($a)=-$sgn($b$), then the statement is technically no longer true, since $|0+0|\not\lt|0|+|0|$ even though sgn($0)=-$sgn($0$).)