Let’s get straight into this one today!

The Triangle Inequality

As the name suggests, this is an equality that can be related back to a triangle. Consider the triangle’s sides below,

The triangle inequality states: the sum of any two side lengths of a triangle is greater than or equal to the third side. In the triangle above the sum of the lengths x and y will have to be larger than or equal to the third side z.

For triangles, this is intuitive. If you had to walk from the top of the triangle to the bottom right, the shortest path would be to travel along z. If we used the language (and notation) of vectors, we would write this as: |\vec{x}+\vec{y}| = | \vec{z}| \leq |\vec{x}| + |\vec{y}|. The same holds true for real numbers too! This captures in the triangle inequality:


Theorem: (The Triangle Inequality) Let x,y \in \mathbb{R}. Then,

|x+y| \leq |x| + |y|.


Comment: This probably seems obvious, I mean |3+7| = |3| + |7|. However, the real fun happens when we have negative numbers e.g. 4=|3-7| \leq |3| + |7| = 10.

Scratch Work: How might we go about this? We will need to use some properties of inequalities. For a refresher,

|x|= \left\{  \begin{array}{lr}       -x & x< 0 \\      \;\;\,x & \;\;x \geq 0.       \end{array}\right.

We’ll we can see that:

Lemma 1: -|x|\leq x\leq |x|.

Proof: We can prove this case by case. When x<0, then -|x|= x\leq |x|. Likewise, when x\geq 0, we have -|x|\leq x= |x|.

\square

Lemma 2: If -a\leq x\leq a for a>0, then |x|\leq a.

Proof: Let -a\leq x\leq a. We do a proof by cases, yet again.

Case 1: x\geq 0. Here, we automatically have x= |x|\leq a from the right-hand side of the inequality.

Case 2: x<0. This case requires a little more work. When x<0 we have x = - |x|. This implies,

-a\leq -|x|<0\leq a

Looking at the left side of the inequality we see,

|x|\leq a

In either case we have |x|\leq a.

\square

We have enough to prove the triangle inequality! See what you can come up with before you continue!

Proof of the Triangle Inequality:

Let x,y \in \mathbb{R}. By lemma 1 we know both,

-|x|\leq x\leq |x| and -|y|\leq y\leq |y|.

Adding these we deduce,

-\left(|x|+|y|\right)\leq x+y\leq |x|+|y|.

Which, by lemma 2 implies the triangle inequality:

\boxed{|x+y| \leq |x| + |y|}.

\square

Closing Remarks

I don’t think it would be a hyperbole if we said that the triangle inequality is the most important inequality for real analysis. Nuf said.

This article was purposely short, so that way the people who only want to learn about the triangle inequality won’t get bogged down with more than they needed! But, for those reading the series in part 2 of this week’s posts we use the triangle inequality to show that when a sequence has a limit, it must be unique! Check it out here!

2 responses to “Let’s Get Real… Analysis (Part 3.1) The Triangle Inequality”

  1. Let’s Get Real… Analysis (Part 3.2) Uniqueness of Limits – A Kick in the Discovery Avatar

    […] and then in part 2 we learned about sequence limits. Part 3 has been broken up into two parts: 3.1 is on the triangle inequality and 3.2 (which is this article) is building on what we did with limits last time. Here, we will […]

    Like

Leave a comment