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I understand intuitively that this is true, but I'm embarrassed to say I'm having a hard time constructing a rigorous proof that $|a+b| \leq |a|+|b|$. Any help would be appreciated :)

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  • $\begingroup$ Isn't this an axiom in metric space? $\endgroup$
    – NECing
    Feb 18, 2013 at 19:11
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    $\begingroup$ There is no addition in metric space. @ShuXiaoLi $\endgroup$
    – k.stm
    Feb 18, 2013 at 19:16
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    $\begingroup$ That a metric must obey the triangle inequality is indeed one of the axioms of a metric space. $\endgroup$
    – user1236
    Jul 28, 2015 at 1:04
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    $\begingroup$ Consider the possibilities for a and b: each can be negative, zero, or positive. Thus there are at most nine possibilities to check out separately. You can do it! Be brave! $\endgroup$ Jan 24, 2018 at 1:18
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    $\begingroup$ The question is not well-defined until you say what $ a $ and $ b $ are: real numbers complex numbers, vectors or something else again. $\endgroup$
    – PJTraill
    Oct 10, 2018 at 20:44

11 Answers 11

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From your definition of the absolute value, establish first $|x| = \max\{x,-x\}$ and $\pm x ≤ |x|$.

Then you can use \begin{align*} a + b &≤ |a| + b ≤ |a| + |b|,\quad\text{and}\\ -a - b &≤ |a| -b ≤ |a| + |b|. \end{align*}

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    $\begingroup$ Clear and concise, +1. $\endgroup$
    – Julien
    Feb 18, 2013 at 19:18
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    $\begingroup$ Nice. Thanks! I got hung up for a sec on this step:$$-a-b\leq|a|-b$$ but then I put in the intermediate step:$$-a-b\leq|-a|-b=|a|-b$$Thank you! $\endgroup$
    – ivan
    Feb 18, 2013 at 19:28
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    $\begingroup$ Do you need that step though? Because $|x|=max\{x,-x\}$, which is trivially greater than or equal to $-x$. $\endgroup$ Mar 8, 2015 at 3:32
  • $\begingroup$ When I use sodiumnitrate's way to prove the two lines, then I don't even need to use the second fact. Because after that we know that $$ \left| a + b \right| = \max\left(a + b, - \left( a + b \right)\right)$$, so no matter which one of $$ \left\{ a + b, -\left( a + b \right) \right\}$$ $ \left| a + b \right|$ is equal to we have that $$\left| a + b \right| \le \left| a \right| + \left| b \right|$$. So is the second fact even necessary? $\endgroup$ Sep 15, 2021 at 2:09
  • $\begingroup$ @cuppajoeman sodiumnitrate “proves” the fact. It’s just a triviality, but having it explicitly stated makes the second inequality line obvious without the need of spending a second thought on it. $\endgroup$
    – k.stm
    Sep 18, 2021 at 16:52
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$$|a|^2+|b|^2+2|a||b|\geq a^2+b^2+2ab$$ $$(|a|+|b|)^2 \geq (a+b)^2\phantom{a}(\because \forall x\in \mathbb{R};\phantom{;}x^2=|x|^2)$$

$$||a|+|b||\geq |a+b|$$

$$\therefore |a|+|b|\geq |a+b|$$

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A simple proof of the triangle inequality that is complete and easy to understand (there are more cases than strictly necessary; however, my goal is clarity, not conciseness).

Prove the triangle inequality $| x | + | y| ≥ | x + y|$.

Without loss of generality, we need only consider the following cases:

  1. $x = 0$
  2. $x > 0, y > 0$
  3. $x > 0, y < 0$

Case $1$. Suppose $x = 0$. Then we have

$| x| = 0$

$| x| + | y| = 0 + | y| = | y|$

Thus $| x| + | y| = | x + y|$.

Case $2$. Suppose $x > 0, y > 0$. Then, since $x + y > 0$, we have

$| x| = x$

$| y| = y$

$| x| + | y| = x + y$

$| x + y| = x + y$

Thus $| x| + | y| = | x + y|$.

Case $3$. Suppose $x < 0, y < 0$. Then, since $x + y < 0$, we have

$| x| = −x$

$| y| = −y$

$| x| + | y| = (−x) + (−y)$

$| x + y| = −(x + y) = (−x) + (−y)$

Thus $| x| + | y| = | x + y|$.

Case $4$. Suppose $x > 0, y < 0$. Then we have

$| x| = x$

$| y| = −y$

$| x| + | y| = x + (−y)$

We must now consider three cases:

a. $x + y = 0$

b. $x + y > 0$

c. $x + y < 0$

Case $4a$. Suppose $x + y = 0$. Then we have

$| x + y | = |0| = 0$

Since $y < 0$, it follows that $−y > 0$ and thus $x + (−y) > 0 + (-y) = -y > 0$.

Therefore, since $| x| + | y| = x + (−y)$, we must have $| x| + | y| > | x + y|$.

Case $4b$. Suppose $x + y > 0$. Then we have

$| x + y| = x + y$

Since $y < 0$, it follows that $−y > 0 > y$ and thus $x + (−y) > x + y$.

Therefore, since $| x| + | y| = x + (−y)$, we must have $| x| + | y| > | x + y|$.

Case $4c$. Suppose $x + y < 0$. Then we have

$| x + y| = −(x + y) = (−x) + (−y)$

Since $x > 0$, it follows that $x > 0 > −x$ and thus $x + (−y) > (−x) + (−y)$.

Therefore, since $| x| + | y| = x + (−y)$, we must have $| x| + | y| > | x + y|$.

This concludes the proof.

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Firstly, observe that $-|x|\leq x\leq|x|$ , and $-|y|\leq y\leq|y|$ follow from the definition of the absolute value function. (Consider the cases $x$ is non-negative and $x$ is negative and what happens to $|x|$, the same goes for $y$ mutatis mutandis).

Since $y\leq|y|$ , then

$$|x|+y\leq|x|+|y|\tag{1}$$

by adding $|x|$ to both sides of $y\leq|y|$ . Likewise, adding $y$ to both sides of $x \leq |x|$ we have: $$y+x \le y+|x|\tag{2}$$

Combining equations $1$ and $2$, and using the transitive property of the relation $\leq$, we have:

$$y+x \le y+|x|\leq |y|+|x|$$ $$y+x\leq |y|+|x|$$

Likewise, since $-|y|\leq y$ , then $-|y|-|x|\leq y-|x|$ by adding $-|x|$ to both sides. But $-|x|\leq x\implies y-|x|\leq y+x$ by adding $y$ to both sides. Thus by the transitive property:

$$ -|y|-|x|\leq y-|x|,\text{ and }y-|x|\leq y+x \implies -|y|-|x|\leq y+x \tag{3}$$

By combining equations 2 and 3, we have:

$$-(|y|+|x|)\leq y+x\leq|y|+|x|$$

Now noting that $|b|\leq a\iff-a\leq b\leq a$ , we have:

$$|x+y|\leq|x|+|y|$$

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    $\begingroup$ Great answer! Thank you for showing all the steps. $\endgroup$
    – Mr X
    Jul 29, 2022 at 23:17
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If a neat algebraic argument does not suggest itself, we can do a crude argument by cases, guided by the examples $a=7,b=4$, $a=-7,b=-4$, $a=7, b=-4$, and $a=-7, b=4$.

If $a\ge 0$ and $b\ge 0$ then $|a+b|=|a|+|b|$.

If $a\le 0$, and $b\le 0$, then $|a+b|=-(a+b)=(-a)+(-b)=|a|+|b|$.

Now we need to examine the cases where $a$ is positive and $b$ is negative, or the other way around. Without loss of generality we may assume that $|b|\le |a|$.

If $a\gt 0$, then $|a+b|=|a|-|b|$. This is $\lt |a|$, and in particular $\lt |a|+|b|$.

If $a\lt 0$, then again $|a+b|=|a|-|b|$.

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The proof given in Wikipedia / Absolute Value is interesting and the technique can be used for complex numbers:

Choose $\epsilon$ from $\{ -1,1\}$ so that $\epsilon (a+b) \ge 0$. Clearly $\epsilon x \le |x|$ for all real $x$ regardless of the value of $\epsilon$, so

$|a+b|= \epsilon (a+b) = \epsilon a + \epsilon b \le |a|+|b|$

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These kind of questions require clarification about the assumptions one begins with. This answer covers the whole package often referred to as "the triangle inequality". I make only the following assumptions, which (with the set $\mathbb R$ of real numbers in place of set $X$) are part of any axiomatization of the real number system:

  • There is a commutative group operation $(a,b) \mapsto a+b$ with neutral element $0$ in the set $X$.
  • There is a linear order $<$ in the set $X$, compatible with $+$ in the sense that if $a<b$, then $a+c<b+c$.
  • The absolute value $|x|\in X$ of $x\in X$ is $-x$ in the case of $x<0$, and $x$ in the cases of $x=0$ and $0<x$.

Note that if $x\neq 0$, then either $x<0$, whence $0<-x=|x|$, or $0<x$, so $0<|x|$ in this case as well. If $|x|=0$, then neither $x<0$ nor $0<x$, since each of these inequalities implies $0<|x|$. These observations prove the following basic result, which is used below, and in which $a\leq b$ stands for "$a<b$ or $a=b$''.

(Positive definiteness.) For each $x\in X$ the inequality $0\leq|x|$ holds, and $0=|x|$ if and only if $x=0$.

Remark. The formal claim $x\leq y$ is also called "inequality". By convention the name "strict inequality" is reserved for the formal claim $x<y$ to underline that it actually claims that the equality $x=y$ does not hold.

The proof of the basic fact that $|-x|=|x|$ holds for every $x\in X$, which I also use below, is immediate:

  1. If $x<0$, then $0<-x$, so $|-x|=-x=|x|$.
  2. If $0<x$, then $-x<0$, so $|-x|=-(-x)=x=|x|$.
  3. If $x=0$, then $-x=x$, so $|-x|=|x|$.

The triangle inequality regarding addition $+$ often refers to the combination of the following two non-strict inequalities together with the statements below them, that clarify when these inequalities are strict:

$$ \big||a|-|b|\big| \leq |a-b| \leq |a|+|b| $$

  • The left inequality is strict if and only if either $a<0<b$ or $b<0<a$.
  • The right inequality is strict if and only if either both $a<0$ and $b<0$, or both $0<a$ and $0<b$.

This more precise formulation of "the triangle inequality" has the following elementary proof by cases:

  1. Case of $0<a$ and $0<b$. In this case clearly $\big||a|-|b|\big|=|a-b|$. Furthermore,

    • If $a<b$, then $a-b<0$, so $|a-b|=-(a-b)=-a+b=-|a|+|b|<0+|b|<|a|+|b|$.
    • If $b<a$, then $0<a-b$, so $|a-b|=a-b=|a|-|b|<|a|+0<|a|+|b|$.
    • If $a=b$, then $a-b=0$, so $|a-b|=0+0<|a|+|b|$.
  2. Case of $a<0$ and $b<0$. Apply case 1 for $-a$ and $-b$ in place of $a$ and $b$, and use $|-x|=|x|$.

  3. Case of $0<a$ and $b<0$. Now $0<a-b$, so $|a-b|=a-b=a+(-b)=|a|+|b|$.

    • If $a<-b$, then $|a|<|b|$, so $|a|-|b|<0$, and therefore $$\big||a|-|b|\big|=-(|a|-|b|)=-(a-(-b))=-a-b<0-b<a-b=|a-b|.$$
    • If $-b<a$, then $|b|<|a|$, so $0<|a|-|b|$, and therefore $$\big||a|-|b|\big|=|a|-|b|=a-(-b)=a+b<a+0<a-b=|a-b|.$$
  4. Case of $a<0$ and $0<b$. Apply case 3 for $-a$ and $-b$ in place of $a$ and $b$, and use $|-x|=|x|$.

  5. Case of $a=0$. In this case $\big||a|-|b|\big|=\big|0-|b|\big|=\big||b|\big|=|b|$, and since $$|b|=|-b|=|0-b|=|a-b| \qquad \text{and} \qquad |b|=0+|b|=|0|+|b|=|a|+|b|,$$ we have $\big||a|-|b|\big|=|a-b|=|a|+|b|$.

  6. Case of $b=0$. Apply 5, switching $a$ with $b$, and use $|-x|=|x|$.

These cases exhaust all the possibilities and are mutually exclusive.

Remark. Instead of the rather cumbersome wording above, the phrases "$a$ and $b$ are of the same sign" and "$a$ and $b$ are of different signs" are commonly used. Then "the sign of $x$" is $+$ in case of $0<x$, and $-$ in case of $x<0$, while the neutral element $0\in X$ is considered as having no sign at all.

I cannot see any significant short-cut in proving the claim above, provided perhaps by some algebra-trick. (There is not much algebra in a linearly ordered commutative group.) We are so accustomed to using triangle inequality while "computing stuff" by applying various algebraic identities, that it is easy to forget that there is logic involved in the triangle inequality. In logic one sometimes just needs to comb through all the cases.

The significance of the triangle inequality is not in some deep insight its proof requires, but rather in its usefulness and in its elegant formulation compared to the tedious case-by-case proof it seems to require.

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I hope that the following proof is the shortest one and make use of the order in $\mathbb{R}$.

  • If $ab\geq 0$ then $$ |a+b|= \begin{cases} a-b & if\quad a\geq 0,\; b\geq 0,\\ -a-b & if\quad a\leq 0,\; b\leq 0 \end{cases} =|a|+|b| $$

  • If $ab<0$ then $$ |a+b|\leq\max\{|a|,|b|\}\leq |a|+|b|. $$

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This proof works alongside the geometric notion that adding numbers on the real line is a 'vector operation'. It also lays out the exact conditions under which the triangle inequality is an equation,

$\quad |x + y| = |x| + |y|$.

Recall that in general,

$\tag 1 a \le b \; \text{ iff } \; \exists \, u \ge 0 \text{ such that } a + u = b$

It is easy to see that whenever $x, y \ge 0$ or $x, y \le 0$ the triangle inequality holds since there is no 'less than' there, $|x+y| = |x| + |y|$.

Logically, there are two case left to handle:

$\quad \text{Case 1: } \left[x \lt 0 \lt y\right] \text{ and } (-x) \le y$

$\quad \text{Case 2: } \left[x \lt 0 \lt y\right] \text{ and } (-x) \ge y$

Since $|z| = |-z|$, it is only necessary to take care of the first case to prove the triangle inequality.

Case 1

Using $\text{(1)}$, we write $(-x) + u = y$ for $u \ge 0$ and $(-x) \gt 0$. Noting that $u \lt y$, we have

$\quad |x + y| = |x + (-x) + u| = |u| = u \lt y \lt (-x) + y = |x| + |y|$

So only when $x$ and $y$ 'straddle $0$' is the triangle inequality a 'strict less than' relation,

$\quad |x + y| \lt |x| + |y|$.

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Since the proof of CW is necessary and similar I will do both

  1. Start both proofs with the fact that a vector dotted with itself is greater than or equal to 0
  2. for CW substitute vector = x-ty, for triangle inequality vector = x+y
  3. for CW, after dotting x-ty with itself let t = (x.y)/(y.y), for triangle ineq. after dotting x+y with itself and getting a quadratic equation with a dot product in the middle, use CW to show that this quadratic is less than or equal to the same quadratic with the moduluses of the vectors of the dot product in the equation on the left
  4. after rearranging both sqrt both sides
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$|x+y|^2=(x+y).(x+y) =(x.x)+2(x.Y)+(y.y) =|x|^2+2(x.Y)+|y|^2$ from Cauchy-Schwarz inequality,$|x.Y|<=|x||y|$ $|x+y|^2 <=|x|^2+2|x||y|+|y|^2$ $ |x+y|^2 <=(|x|+|y|)^2$ taking square root on both sides. $|x+y|<=(|x|+|y|)$

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    $\begingroup$ Welcome to stackexchange. That said, answering a five year old question with many good answers is not necessarily a good use of your time or ours (since the answer bumps the question to "recently active" so lots of folks look at it. Do pay attention to new questions where you can help. And do use mathjax to format mathematics: math.meta.stackexchange.com/questions/5020/… $\endgroup$ Dec 27, 2018 at 1:23

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