Yet another proof of the triangle inequality.

In this article, we're proving the triangle inequality, which can be understood as follows:

The shortest distance between two points is a straight line.

I like this theorem because it feels intuitive ... like it needs to be true.

So let's state this theorem in mathematical language. For any two vectors $\vec{u},\vec{v}\in\R^n$, the length of their sum vector (the direct path) is always shorter than the sum of their indivdual path lengths (the detour):

Here, $||\cdot|| : \R^n \to \R$ denotes a vector's norm, defined by $||\vec{x}|| = \sqrt{\vec{x} \cdot \vec{x}}$, where we use the $\vec{x}\cdot\vec{y}$ notation for the dot product $:\R^n\times\R^n\to\R$ of two vectors.

Lemma: The Cauchy-Schwarz Inequality

First we'll be proving that the absolute value of a dot product is less than the product of the individual vector's norms, which is known as the Cauchy-Schwarz inequality:

Given any two vectors $\vec{u}, \vec{v} \in \R^n$, let us define $\vec{w} = \vec{u} - t\vec{v}$, for some arbitrary scalar $t\in\R$. By distributivity and scalar multiplication of the dot product, we have

We're about to use the fact that $|a + b| \leq |a| + |b|$ for any $a,b\in\R$.

Why is this true? Note how $|a|\geq a$ and $|b|\geq b$, which implies that $|a|\cdot|b|\geq a\cdot b$. It follows that $a^2 + 2|a||b| + b^2 \geq a^2 + 2ab + b^2$, implying that $(|a| + |b|)^2 \geq (a + b)^2$, from which it follows that $|a| + |b|\geq |a + b|$

Continuing from our previous point:

And, since $||\vec{w}||^2 \geq 0$, it follows that

The key observation here, is that this is a quadratic inequality in the variable $t$. Geometrically speaking, this inequality states that a parabola, parameterized by $t$, should cross the x-axis at most once.

Since the well known quadratic formula tells us that a function $f(x) = ax^2 + bx + c$ intersects with the x-axis at coordinates

it follows that, for the above inequality to hold, we must have $b^2 - 4ac \leq 0$, i.e.:

By rearranging some terms, we find that $(\vec{u}\cdot\vec{v})^2 \leq (||\vec{u}||\cdot||\vec{v}||)^2$, from which the Cauchy-Schwarz inequality follows.

Proof for The Triangle Inequality

By using the Cauchy-Schwarz inequality, the proof for the triangle inequality becomes relatively straightforward.

Let $\vec{u}$ and $\vec{v}$ be arbitrary vectors in $\R^n$. Then:

From this, the triangle inequality follows: $||\vec{u} + \vec{v}|| \leq ||\vec{u}|| + ||\vec{v}||$