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Proof of Triangle Inequality

The triangle inequality is a very important mathematical relation. It is a core requirement for metrics, or distance functions, to be considered valid and furthermore, a set is equipped with a metric and is therefore considered a metric space if there exists a distance function defined for all members of the set that satisfies the triangle inequality, along with two other requirements that are usually fairly easy to satisfy. Here, we give a proof for the triangle inequality in \(\mathbb{R}^{k}\), which is fairly restrictive, as the inequality can of course be generalized to include all metric spaces. It is assumed in the following proof that the reader is familiar with the concepts of a Euclidean norm (aka magnitude of a vector) and Euclidean inner product (aka dot product of two vectors).

The triangle inequality in \(\mathbb{R}^{k}\) is: \(\parallel \vec{x} +\vec{y} \parallel \leq \parallel \vec{x} \parallel + \parallel \vec{y} \parallel \) \(\forall \vec{x}\), \(\vec{y}\) \(\in \mathbb{R}^{k}\)


Consider the Pythagorean Theorem: \(a^2 +b^2 =c^2\). Now, let \(a\) and \(b\) be the Euclidean norms of two vectors in \(\mathbb{R}^{k}\), \(\vec{x}\) and \(\vec{y}\), respectively. Then it follows that \(c\) is the Euclidean norm of \( \vec{x} +\vec{y} \), due to the geometric properties of vector addition. Then we have:

\(\parallel \vec{x} +\vec{y} \parallel^2 = \parallel \vec{x} \parallel^2 + \parallel \vec{y} \parallel^2 \)

This is equivalent to stating: \((\vec{x} +\vec{y})\cdot(\vec{x} +\vec{y}) = \vec{x}\cdot\vec{x} + \vec{y}\cdot\vec{y}\)

Now let \(\vec{x} = <x_1,x_2,...,x_k>\) and \(\vec{y} = <y_1,y_2,...,y_k>\)

Then it is apparent that: \(\vec{x}+\vec{y} = <x_1+y_1,x_2+y_2,...,x_k+y_k>\)

And: \((\vec{x} +\vec{y})\cdot(\vec{x} +\vec{y}) = \displaystyle \sum_{i=1}^{k} (x_i+y_i)^2 \)

\(\vec{x}\cdot\vec{x}=\displaystyle \sum_{i=1}^{k} x_i^2\)

\(\vec{y}\cdot\vec{y}=\displaystyle \sum_{i=1}^{k} y_i^2\)

Then we have: \(\displaystyle \sum_{i=1}^{k} (x_i+y_i)^2=\displaystyle \sum_{i=1}^{k} x_i^2 + \displaystyle \sum_{i=1}^{k} y_i^2\)

\(\Rightarrow\) \(\sqrt{\displaystyle \sum_{i=1}^{k} (x_i+y_i)^2}=\sqrt{\displaystyle \sum_{i=1}^{k} x_i^2 + \displaystyle \sum_{i=1}^{k} y_i^2}\)

Now, it is clear that:

\(0 \leq 2\sqrt{(\displaystyle \sum_{i=1}^{k} x_i^2)(\displaystyle \sum_{i=1}^{k} y_i^2)}\)

\(\Rightarrow\) \(\displaystyle \sum_{i=1}^{k} x_i^2 + \displaystyle \sum_{i=1}^{k} y_i^2 \leq \displaystyle \sum_{i=1}^{k} x_i^2 + \displaystyle \sum_{i=1}^{k} y_i^2 +2\sqrt{(\displaystyle \sum_{i=1}^{k} x_i^2)(\displaystyle \sum_{i=1}^{k} y_i^2)}\)

Note: \(\displaystyle \sum_{i=1}^{k} x_i^2 + \displaystyle \sum_{i=1}^{k} y_i^2 +2\sqrt{(\displaystyle \sum_{i=1}^{k} x_i^2)(\displaystyle \sum_{i=1}^{k} y_i^2)}= \left(\sqrt{\displaystyle \sum_{i=1}^{k} x_i^2}+\sqrt{\displaystyle \sum_{i=1}^{k} y_i^2}\right)^2\)

Then: \(\displaystyle \sum_{i=1}^{k} x_i^2 + \displaystyle \sum_{i=1}^{k} y_i^2 \leq\left(\sqrt{\displaystyle \sum_{i=1}^{k} x_i^2}+\sqrt{\displaystyle \sum_{i=1}^{k} y_i^2}\right)^2\)

\(\Rightarrow\) \(\sqrt{\displaystyle \sum_{i=1}^{k} x_i^2 + \displaystyle \sum_{i=1}^{k} y_i^2} \leq\left(\sqrt{\displaystyle \sum_{i=1}^{k} x_i^2}+\sqrt{\displaystyle \sum_{i=1}^{k} y_i^2}\right)\)

Since: \(\sqrt{\displaystyle \sum_{i=1}^{k} (x_i+y_i)^2}=\sqrt{\displaystyle \sum_{i=1}^{k} x_i^2 + \displaystyle \sum_{i=1}^{k} y_i^2}\)

We have: \(\sqrt{\displaystyle \sum_{i=1}^{k} (x_i+y_i)^2}\leq\left(\sqrt{\displaystyle \sum_{i=1}^{k} x_i^2}+\sqrt{\displaystyle \sum_{i=1}^{k} y_i^2}\right)\)

\(\Rightarrow\) \(\sqrt{(\vec{x} +\vec{y})\cdot(\vec{x} +\vec{y})} \leq \sqrt{\vec{x}\cdot\vec{x}} +\sqrt{ \vec{y}\cdot\vec{y}}\)

\(\Rightarrow\) \(\parallel \vec{x} +\vec{y} \parallel \leq \parallel \vec{x} \parallel + \parallel \vec{y} \parallel \)

Which is the triangle inequality.


Remark: Some may be wondering why we were able use the Pythagorean Theorem in generalized \(\mathbb {R}^{k}\) without restricting \(k\) to \(k=2\). This is because in higher dimensions of Euclidean Space, any triangle existing in this metric space can be considered a flat projection on some arbitrary 2-D plane in Euclidean k-space. Since all planes in such space are subject to the same metric and inner product as the space itself, this usage is justified.

Note by Ethan Robinett
3 years, 2 months ago

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Can you add this to the Triangle Inequality Wiki page? I also like several of your other notes, can you look into moving them into the Wiki? Thanks! @Ethan Robinett

Calvin Lin Staff - 3 years ago

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a²/x - b²/y= 0 a²b/x + b²a/y=a+b

Hitesh Hitesh - 3 years, 1 month ago

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