Show that the jungle river (barbed wire) metric is a metric I want to prove that Jungle River metric is indeed a metric space, and determine it is open and closed balls. Firstly, i know that the metric is given by $x,y\in \mathbb{R}^2$, such that $x=(x_1,x_2), y=(y_1,y_2),$ and $$
d(x,y)
= \begin{cases}
|x_2-y_2|, & \text{if } x_1 = y_1,\\
|x_2| + |y_2| + |x_1-y_1|, & \text{if } x_1 \neq y_1.
\end{cases}$$
 A: If $d(x,y)=0,$ then $x_1=x_2$ (why?) and therefore  $|y_1-y_2|=0$ which then yields $y_1=y_2.$ Also, if $x=y,$ then $(x_1,x_2)=(y_1,y_2),$ so that $$d(x,y)=|x_2-y_2|=0.$$
To show that $d(x,y)\leq d(x,z)+d(z,y), z:=(z_1,z_2)$; Consider the following four cases: (a) $x_1=z_1;$ (b) $x_1,x_2,z_1$ are pairwise different; (c) $x_1=x_2$ but $z_1$ is different  (d) $x_1$ is different from $$x_2=z_1.$$
Note that: Drawing some balls in this metric you just seem to get a diamonds of some size along the x-axis and then potentially a vertical line extending from its centre that passes through the tip.  
A: I'm also struggling a bit with proving the triangle inequality with the cases proving in @Yusuf's answer and hope to revive this post a little by positing what I've got so far:
Let $x,y,z \in \mathbb{R}^2$.
Case 1: $x_1 = z_1$.
Then, if (1a) $x_1 \neq y_1$ and $y_1 \neq z_1$ we have
\begin{align*}
 d(x,z)
& = | x_2 - z_2 |
\le | x_2 | + | z_2 | + | x_1 - z_1 | \\
& \le | x_2 | + | z_2 | + 2 | y_2 | +  | x_1 - y_1 + y_1 - z_1 | \\
& \le | x_2 | + | y_2 | + | x_1 - y_1 | + | z_2 | +  | y_2 | + | y_1 - z_1 |
= d(x,y) + d(y,z).
\end{align*}
and if (1b) $x_1 = y_1 = z_1$
\begin{align*}
d(x,z)
= | x_2 - z_2 |
\le |x_2 - y_2 + y_2 - z_2 |
\le |x_2 - y_2 | +| y_2 - z_2 |.
\end{align*}
Case 2: $x_1 \neq x_2 \neq z_1$.
Then for (2a) $y_1 \neq z_1$ we have
\begin{equation*}
d(x,z)
= | x_2 | + | z_2 | + | x_1 - z_1 |
\overset{\triangle \neq }{\le} | x_2 | + |y_2 | + | x_1 - y_1 | + | y_2 | + | z_2 | + | y_1 - z_1 |
\end{equation*}
For (2b) $y_1 = z_1 \neq x_2$ we would have to show
\begin{equation*}
d(x,z)
= | x_2 | + | z_2 | + | x_1 - z_1 |
\le | x_2 | + | y_2 | + | x_1 - y_1 | + | z_2 - y_2 |,
\end{equation*}
which I couldn't manage.
Case 3: $x_1 = x_2 \neq z_1$
If also $z_1 \neq y_1$ we have Case (1a).
For $z_1 = y_1$ we need to show
\begin{align*}
d(x,z)
& = | x_2 | + | z_2 | + | x_1 - z_1 |
= | x_2 | + | z_2 | + | x_2 - z_1 | \\
& \overset{(\star)}{\le} | x_2 | + | y_2 | + | x_2 - y_1| + | y_2 - z_2 | \\
& = | x_2 | + | y_2 | + | x_1 - y_1| + | y_2 - z_2 |,
\end{align*}
but I haven't been able to show $(\star)$, which is very similar to (2b)
Case 4: $x_1 \neq x_2 = z_1$.
For (4a) $y_1 = z_1$ we would have to show
\begin{align*}
d(x,z)
& = | x_2 | + | z_2 | + | x_1 - z_1 |
= | z_1 | + | z_2 | + | x_1 - z_1 | \\
& \overset{(\star)}{\le} | z_1 | + | y_2 | + | x_1 - y_1 | + |y_2 - z_2|,
\end{align*}
but I haven't been able to show $(\star)$, which is very similar to (2b)
