Prove that it is a metric space my uni is closed because of the pandemy and I'm home-learning metric spaces. I can't solve one problem from our book.

On set $R_m \times R_m$ we define a function $d_3$, so that if
  $x=[x_1, x_2, ..., x_m]$, $y=[y_1, y_2, ..., y_m]\in R_m$, we set
  $d_3(x,y)=0$ if $x=y$ (that's when $x_j=y_j$ for every $j=1,2,...,m$).
  And when $x\neq y$, then $d_3(x,y)=\frac{1}{s}$, where $s$ is the
  smallest of numbers $n \leq m$, for which $x_n \neq y_n$
Prove that $d_3$ is a metric space on $R_m$.

I know how to prove symetry and identity, but I don't know how to prove triangle inequality. 
Could you please help me?

The triangle inequality for a measure says that given $x,y,z$ in our set, it is always true that $d(x,y) \leq d(x,z)+d(y,z)$. 
In this setting, for $d_3$, we can just look at the 4 possible cases:
Case 1) $x=z$ and $z=y: d_3(x,z)+d_3(y,z)=0=d_3(x,y) \checkmark$
Case 2) $x \neq z$ and $z=y: d_3(x,z)+d_3(y,z)=\frac{1}{s_{xz}}+0 = \frac{1}{s_{xy}} = d_3(x,y) \checkmark$
Case 3) $x=z$ and $z \neq y: d_3(x,z)+d_3(y,z)=0+\frac{1}{s_{yz}} = \frac{1}{s_{xy}} = d_3(x,y) \checkmark$
Case 4) $x \neq z$ and $z \neq y: d_3(x,z)+d_3(y,z)=\frac{1}{s_{xz}}+\frac{1}{s_{yz}} > \frac{1}{s_{xy}} = d_3(x,y) \checkmark$
I think it should be clear in the first 3 cases why the $s$ are equal. The interesting part for me is the 4th case - thinking about why the inequality holds. I'll leave that to you.
