Proof that the Hamming distance is a metric [duplicate]

If $x$, $y$, are words of length $n$ in a code $C$, $x=x_1 \cdots x_n$, $y=y_1 \cdots y_n$ we define

$$d(x, y)= d(x_1, y_1) + \cdots + d(x_n, y_n)$$

from where

$$\quad d(x_i, y_i)=0 \text{ if } x_i=y_i \quad \text{ and } d(x_i, y_i)=1 \text{ if } x_i \neq y_i$$

So let's review the characteristics that a metric meets.

1. My idea to proof for non-negativity is to separate the summands, on the one hand those that cancel out and on the other hand those that remain distinct, i.e.

$$d(x, y)=\sum_{i=1, x_i \neq y_i}^q d(x_i, y_i)+ \sum_{i=1, x_i = y_i}^p d(x_i, y_i)= (1+ \cdots +1) + 0 \geq 0$$

2. Lets check out that $d(x, y)=0$ iff $x=y$If$\sum_{i=1}^n d(x_i, y_i)=0$, as $d(x_i, y_i) \geq 0$, $d(x_i, y_i)=0$, thus $x_i=y_i$ and $x=y$The converse is similar

3)For the symmetry I did something similar to side 1), split the sum in two and commute the resulting summands.

Now, for the triangular inequality I have tried to see it by cases, but I don't see it at all clear, any help? I appreciate it