# How can one lone picture prove the Triangle Inequality, $|x−y|≤|x|+|y|$, $|x|−|y|≤|x−y|$, and the Reverse Triangle Inequality?

I always showcase separate pictures of Triangle Inequality, and Reverse, to 16-years-old students in 1st class. I reshow pictures in 2nd class. I preachify

Please remember these 4 inequalities. Save these pictures as screen savers or wallpapers, hang them by your toilet and bedroom, do what you got to do. In my 10 years of teaching, someone YEARLY always forget 1 of below 4 inequalities ON MULTIPLE TESTS! Do not be this addlepate!

But no soap! Yearly, someone still forgets! After they lost marks, they and I always scratch our heads! They wonder why they forgot. And I wonder why, after ten years of teaching, I again fail to drum all these 4 inequalities into ALL students' heads. I need new approaches! Rather than separate picture for each inequality, how can I pictorialize below 4 inequalities concurrently?

#### How can 1 sole picture prove all 4 inequalities at once? Please contrast $$\vec{x} ≠ \vec{y}$$, and draw $$|\vec{x} - \vec{y}| \neq |\vec{x} + \vec{y}|$$ with different lengths.

The four inequalities to be remembered are the ones in the following excerpts from Spivak's Calculus Fourth Edition (2008):

Page 12 proved $$|x + y| ≤ \color{YellowGreen}{|x| + |y|}$$.

Page 16, exercise 12.

(iv) $$\color{red}{|x−y|}≤\color{YellowGreen}{|x|+|y|}$$. (Give a very short proof.)
(v) $$\color{deepskyblue}{|x|−|y|}≤\color{red}{|x−y|}$$. (A very short proof is possible, if you write things in the right way).

(vi) $$| \; \color{deepskyblue}{|x|−|y|} \; |≤|x−y|$$ (Why does this follow immediately from (v)?)

I shalln't blockquote this (vi), the Reverse △ ≠, as I deleted Spivak's superfluous round brackets.

• I remember seeing this question before. Have you changed your name? Commented Oct 22, 2023 at 6:58
• I learned the triangle inequality as $|x + y| \le |x| + |y|$, with addition instead of subtraction. It's equivalent to your presentation by just changing the sign of $y$, but I wonder if it might be easier to remember since it doesn't appear to treat $x$ and $y$ differently (by subtracting one from the other). That is, unless you're teaching a course where the assessment will require students to state the exact version with subtraction. Commented Oct 22, 2023 at 12:48
• @kaya3 "you're teaching a course where the assessment will require students to state the exact version with subtraction." Yes, my students are required to know all these 4 inequalities, even though they are equivalent. They shall be assessed on all 4. Commented Oct 22, 2023 at 21:31
• @kaya3 "When I first read this, I thought you meant the students forgot the formulas of the inequalities." Correct. "But the quoted text suggests to me that you're asking the students to prove the inequalities, and every year, someone forgets one of the proofs." Not merely the proofs, but they also forget at least 1 of these 4 inequalities! These 4 inequalities are required, and ubiquitary, in International Baccalaureate (IB) math. Even if a question doesn't test these inequalities outright, students are required to recall and apply them on their own initiative. Commented Oct 22, 2023 at 21:43
• "I always showcase separate pictures of Triangle Inequality, and Reverse, to 16-years-old students in 1st class. I reshow pictures in 2nd class." Instead of showing pictures for the finished inequalities, have you tried using pictures that show how the four forms of what is essentially one inequality relate to each other? Or would that not accomplish your goal? Commented Oct 23, 2023 at 16:33

Another possibility is to show a sequence of diagrams explaining how the inequalities relate to one another.

The top left diagram shows the triangle inequality, $$\lvert \mathbf{a}+\mathbf{b}\rvert \le \vert\mathbf{a}\rvert + \lvert\mathbf{b}\rvert$$.

The top right diagram shows that $$-\mathbf{b}$$ is just as good a vector as $$\mathbf{b}$$, implying that the triangle $$(\mathbf{a},-\mathbf{b},\mathbf{a}-\mathbf{b})$$ satisfies the inequality just as much as the triangle $$(\mathbf{a},\mathbf{b},\mathbf{a}+\mathbf{b})$$ does. This gives $$\lvert \mathbf{a}-\mathbf{b}\rvert \le \vert\mathbf{a}\rvert + \lvert\mathbf{b}\rvert$$.

Returning, in the bottom left, to the original triangle, with the sum $$\mathbf{a}+\mathbf{b}$$ denoted by $$\mathbf{s}$$, we see that $$\lvert \mathbf{s}\rvert \le \vert\mathbf{a}\rvert + \lvert\mathbf{s}-\mathbf{a}\rvert$$ or $$\lvert \mathbf{s}\rvert - \vert\mathbf{a}\rvert \le \lvert\mathbf{s}-\mathbf{a}\rvert$$.

In the bottom right diagram, the roles of $$\mathbf{a}$$ and $$\mathbf{s}$$ are swapped (which reverses the direction of the red vector), showing that $$\lvert \mathbf{a}\rvert - \vert\mathbf{s}\rvert \le \lvert\mathbf{a}-\mathbf{s}\rvert = \lvert\mathbf{s}-\mathbf{a}\rvert$$. Since at least one of $$\lvert\mathbf{s}\rvert-\lvert\mathbf{a}\rvert$$ and $$\lvert\mathbf{a}\rvert-\lvert\mathbf{s}\rvert$$ is nonnegative, $$\lvert\mathbf{s}-\mathbf{a}\rvert \ge \lvert \lvert \mathbf{s}\rvert - \vert\mathbf{a}\rvert \rvert$$.

In the comments there there was a request for a way of seeing the inequality in the fourth diagram, $$\lvert \mathbf{a}\rvert - \vert\mathbf{s}\rvert \le \lvert\mathbf{a}-\mathbf{s}\rvert$$, "without algebra". The following visualization helps me. Here two sides of the triangle gradually "collapse" onto the third side, with $$\lvert\mathbf{a}-\mathbf{s}\rvert$$ clearly covering $$\lvert \mathbf{a}\rvert - \vert\mathbf{s}\rvert$$ in the final position.

The inequality in the third diagram is evident because $$\lvert\mathbf{s}\rvert - \vert\mathbf{a}\rvert$$ is negative.

A different visualization of the four inequalities is obtained by drawing the usual parallelogram showing vector addition, that is, the parallelogram two of whose opposite sides are copies of $$\mathbf{a}$$, the other two, copies of $$\mathbf{b}$$. One diagonal of this parallelogram is $$\mathbf{a}+\mathbf{b}$$; the other is $$\mathbf{a}-\mathbf{b}$$ (or $$\mathbf{b}-\mathbf{a}$$ if the arrow is reversed). Each diagonal divides the parallelogram into two congruent triangles. For any of these four triangles, the sum of lengths of two sides equals or exceeds the length of the third. This produces six distinct inequalities. The two omitted from your post are $$\lvert\mathbf{a}+\mathbf{b}\rvert\ge\lvert\mathbf{a}\rvert-\lvert\mathbf{b}\rvert$$ and $$\lvert\mathbf{a}+\mathbf{b}\rvert\ge\lvert\mathbf{b}\rvert-\lvert\mathbf{a}\rvert$$, which, together, give $$\lvert\mathbf{a}+\mathbf{b}\rvert\ge\lvert \lvert\mathbf{a}\rvert-\lvert\mathbf{b}\rvert \rvert$$.

A sample of the sagemath code used to generate these diagrams is below.

topleft = dict(horizontal_alignment='left', vertical_alignment='top')
topright = dict(horizontal_alignment='right', vertical_alignment='top')
botleft = dict(horizontal_alignment='left', vertical_alignment='bottom')
botright = dict(horizontal_alignment='right', vertical_alignment='bottom')

a = vector([1, 4])
b = vector([-3, -1])
plot2 = (plot(a, color='blue') +
plot(b, rgbcolor=(1,0.8,0.8), start=a) +
plot(a + b, rgbcolor=(.8,.8,.8)) +
plot(-b, color='red', start=a) + plot(a - b, color='black') +
plot(point((1, 4), size=52, color='red')) +
text(r'$$\mathbf{a}$$', (0.5, 2), **topleft, fontsize=12, color='blue') +
text(r'$$\mathbf{-b}$$', (2.5, 4.5), **topleft, fontsize=12, color='red') +
text(r'$$\mathbf{a-b}$$', (2, 2.5), **topleft, fontsize=12, color='black') +
text(r'$$\mathbf{b}$$', (-0.5, 3.4), **topleft, fontsize=12, rgbcolor=(1,0.8,0.8)) +
text(r'$$\mathbf{a+b}$$', (-1, 1.5), **topright, fontsize=12, rgbcolor=(0.8,0.8,0.8)))
plot2.show(aspect_ratio=1, xmin=-2, xmax=4, ymin=0, ymax=5)


• Thanks so much! Awesome pictures! (1) What program did you use? (2) Have you considered drawing g like this picture, to clarify and underline that $|a - \color{yellowgreen}{g}| \lneq |a| + |\color{greenyellow}{g}|$? (3) How can you pictorialize $\lvert \mathbf{s}\rvert - \vert\mathbf{a}\rvert \le \lvert\mathbf{s}-\mathbf{a}\rvert$ in one go? Your picture doesn't directly prove $|s| - |a| ≤ |s - a|$, but requires 1 step of algebra. Commented Oct 28, 2023 at 0:44
• (1) I used sagemath. I can provide samples of code if you're interested. (2) Can you clarify? If $\mathbf{g}$ is a scalar multiple of $\mathbf{a}$ and if the scalar is nonpositive, the the equality does hold. (3) This is a good question. You can also ask why the top left picture proves the original triangle inequality. For me, turning geometric intuition into an algebraic proof is not straightforward, but nevertheless I find the geometric intuition helpful. If $\lvert\mathbf{a}\rvert\ge\lvert\mathbf{s}\rvert$, the inequality is obvious, since $\lvert\mathbf{s}-\mathbf{a}\rvert$ can't be... Commented Oct 28, 2023 at 3:12
• ...negative. If $\lvert\mathbf{s}\rvert>\lvert\mathbf{a}\rvert$, the my intuition comes from imagining the mechanical procedure of collapsing the triangle (while keeping the side lengths fixed). I''ll try to come up with a visualization of this. Commented Oct 28, 2023 at 4:28
• (1) Thanks again for the animation, and parallelogram! Yes, please provide the code! (2) I meant that my picture, and the $\color{yellowgreen}{\mathbf{g}}$, pictorializes $|a - \color{yellowgreen}{g}| \lll |a| - |\color{greenyellow}{g}|$ more strikingly. Do you see why? Your $\color{dodgerblue}{a}$ and $\color{red}{b}$ are merely a little longer than $a - b$. But in mine, $\color{yellowgreen}{\mathbf{g}}$ is almost the opposite direction of $\color{royalblue}{a}$. Commented Oct 28, 2023 at 18:39
• I added some code. The animation was done with the "animate" command. (Provide it with a list of still images.) Commented Oct 29, 2023 at 15:27

First of all, you want them them to "get it right" on a test, rather than be able to use this to more deeply understand vectors. In fact, it seems to be about memorizing this, which is totally inappropriate for something like this. Please approach math with the goal of sharing its beauty and depth. (Not "no addlepates".)

I recommend that you stop trying to make one perfect handout, and start thinking about how your students learn. We learn through play. Make a handout that encourages them to play. Saving your pictures is not going to help. Walk them through the details on a few problems, and then have them do other problems. Maybe have them make up their own vectors.

Next, use a notation that helps students discuss this. There are x and y coordinates. Please don't make the vectors x and y also. Let's use a and b for vectors. Then $$\left | a \right | = \sqrt{{a_{x}}^{2}+{a_{y}}^{2}}$$. etc.

(If you are teaching 16-year-olds, why are you quoting Spivak?)

• "In fact, it seems to be about memorizing this, which is totally inappropriate for something like this. Please approach math with the goal of sharing its beauty and depth." I agree wholly! See my comments above to @kaya3." Commented Oct 22, 2023 at 21:56
• "I recommend that you stop trying to make one perfect handout, and start thinking about how your students learn. We learn through play. Make a handout that encourages them to play. Saving your pictures is not going to help." I agree that 4 DIFFERENT pictures (1 pic for each inequality) can fail. This is why I need 1 picture showing all 4 inequalities at once, which I've never tried or shown to my students. I want to try this 4-inequalities-in-1 picture first.. Commented Oct 22, 2023 at 21:58
• @Raciquel "(If you are teaching 16-year-olds, why are you quoting Spivak?)" Because these 4 inequalities also appear in other calculus textbooks, but I don't have them at on hand. I merely have Spivak at home, when I posted this. Sorry if I addled you. For example, (iv) above is Exercise 52 on page 17 of Calculus: One and Several Variables by Salas, Etgen, Hille. (v) is Exercise 57 on page 52. (vi) is Exercise 53 on page 17, ibid. Commented Oct 22, 2023 at 21:58

Please remember these 4 inequalities. Save these pictures as screen savers or wallpapers, hang them by your toilet and bedroom, do what you got to do.

I always showcase separate pictures . . . I wonder why, after ten years of teaching, I again fail to drum all these 4 inequalities into ALL students' heads. I need new approaches! Rather than separate picture for each inequality, how can I pictorialize below 4 inequalities concurrently?

Your goal is for students to remember these inequalities, and you're showing pictures because pictures generally make things easier to remember (not because you want the students to prove the inequalities).

However, in this case, pictures are not the best memory aid. You have to remember the pictures and think critically about them in order to translate them into symbols. Critical thinking is always going to be a struggle for some students.

The easiest way to remember these inequalities is to combine them into a single compound inequality that looks so remarkably intuitive that it just "feels" right and is easy to reconstruct based on simple patterns, without having to engage in critical thinking.

\begin{align*} \begin{matrix} \\ |x| - |y|, \\ | \,\, |x| - |y| \,\, | \phantom{,} \end{matrix} \, \leq \, \begin{matrix} \\ |x+y|, \\ |x-y| \phantom{,} \\ \end{matrix} \, \leq |x| + |y| \\ \\ \end{align*}

Here's how to memorize the inequality:

1. Put $$\,\, \begin{matrix} |x+y|, \\ |x-y| \phantom{,} \\ \end{matrix} \,\,$$ in the middle.

2. "Distribute" the absolute value to get the top LHS and RHS. (LHS becomes "-" because left is negative direction; RHS changes to "+" because right is positive direction.) \begin{align*} |x| - |y| \, \leq \, \begin{matrix} \\ |x+y|, \\ |x-y| \phantom{,} \\ \end{matrix} \, \leq |x| + |y| \\ \\ \end{align*}

3. If you put another set of absolute value bars around the LHS, the inequality also holds true. (You can do this to the RHS too; it's just not noteworthy because it's mathematically equivalent to the existing RHS.) \begin{align*} \begin{matrix} \\ |x| - |y|, \\ | \,\, |x| - |y| \,\, | \phantom{,} \end{matrix} \, \leq \, \begin{matrix} \\ |x+y|, \\ |x-y| \phantom{,} \\ \end{matrix} \, \leq |x| + |y| \\ \\ \end{align*}

You are telling sixteen-year olds that:

$$|x - y| \le |x| + |y|$$

... and you call that "triangle inequality"?

... and your teenage students don't get that?

Well, I'm a mathematician and I have troubles understanding that.

However, let's look at this in the original way (based on vectors):

$$|\vec{x} - \vec{y}| \le |\vec{x}| + |\vec{y}|$$

Well: this makes perfectly sense, as you can see from this easy drawing (made in Geogebra, I don't know how to put arrows on the labels of $$u$$ and $$v$$):

The distance from $$(0,0)$$ to $$(8,0)$$ ($$|\vec{u} + \vec{v}|$$) is larger than the distance from $$(0,0)$$ to $$(2,1)$$ ($$\vec{u}$$) plus the distance from $$(2,1)$$ to $$(8,0)$$ ($$\vec{v}$$), which is quite obvious because the smallest distance between two points is a straight line.

Now: set $$\vec{u}=\vec{x}-\vec{y}$$ and $$\vec{v}=\vec{y}$$, and you have your formula in vector-form.
Next you just use 1D-vectors and you have your formula in number-form.

• No. I am calling |x + y| ≤ |x| + |y| Triangle Inequality, as usual. Your picture does not answer my question, because you pictorialized |x + y| ≤ |x| + |y|, then changed variable. Thus, you took 2 steps to prove |x $\color{red}{-}$ y| ≤ |x| + |y|. But I need 1 picture that directly shows |x $\color{red}{-}$ y| ≤ |x| + |y|, without change of variable. Commented Oct 23, 2023 at 8:54
• You can write tex in the "caption" section of the vector but not the "name". You can then choose to label the object using "caption". Commented Oct 23, 2023 at 9:24