I’ll follow the guided steps for this proof. First, I’ll write \(|u+v|^2\) as a dot product. I use the fact that the square of the length of a vector is the same as the dot product of a vector with itself.
\begin{equation*}
|u+v|^2 = (u+v) \cdot (u+v)
\end{equation*}
The next step is to distribute the right side. The dot product distributes over vector addition like ordinary multipliction. Since the dot product is commutative, \(u
\cdot v\) is the same as \(v \cdot u\text{,}\) so I can combine the two term when I distribute the dot product.
\begin{equation*}
|u+v|^2 = u \cdot u + 2 u \cdot v + v \cdot v
\end{equation*}
The middle term has the dot product which I can replace as instructed in the third step.
\begin{equation*}
|u+v|^2 = u \cdot u + 2 |u||v| \cos \theta + v \cdot v
\end{equation*}
The cosine term is at most \(1\text{,}\) so the maximum for the right side happens when I replace the cosine with 1.
\begin{equation*}
|u+v|^2 \leq u \cdot u + 2 |u||v| + v \cdot v
\end{equation*}
Then I need to turn the right side into a square term. The first step in that direction is to write the dot products as squares of lengths.
\begin{equation*}
|u+v|^2 \leq |u|^2 + 2 |u||v| + |v|^2
\end{equation*}
Then the right side is the square of a binomial.
\begin{equation*}
|u+v|^2 \leq \left( |u| + |v| \right)^2
\end{equation*}
Finally, I take the square root of both sides. The square root of positive numbers preserves inequalities (\(a \leq
b\) if an only if \(\sqrt{a}\leq \sqrt{b}\)), so applying the square root is a valid operations for sovling an inequality like this.
\begin{equation*}
|u+v| \leq |u| + |v|
\end{equation*}
I have produced the triangle inequality.
