Show that $(\textbf{R}^{n},d_{1})$ is indeed a metric space.
My solution
The function $d_{1}$ is indeed positive-definite and symmetric on the variables $x$ and $y$. It remains to prove it satisfies the triangle inequality.
To begin with the demonstration, let us consider $a,b\in\textbf{R}^{n}$. Then we have that \begin{align*} |a_{1}| + \ldots + |a_{n}| + |b_{1}| + \ldots + |b_{n}| & = (|a_{1}| + |b_{1}|) + \ldots + (|a_{n}| + |b_{n}|)\\\\ & \geq |a_{1} + b_{1}| + \ldots + |a_{n} + b_{n}| \end{align*}
Consequently, if we take $a = x - y$ and $b = y - z$, we conclude that $d(x,y) + d(y,z)\geq d(x,z)$.
Am I doing it correctly? Is there another way to solve this problem?
Edit
In this context, we define $d_{1}(x,y)$ as follows \begin{align*} d(x,y) = |x_{1} - y_{1}| + \ldots + |x_{n} - y_{n}| \end{align*}