<s>[INST] Proving $d_1$ is a metric. If $d$ is a metric on a set X, then $d_1 = \frac{d(x,y)}{1+d(x,y)}$ is also a metric.
I have proved the other conditions of being a metric except the triangle inequality. Please help!!
 [/INST] We have
$$\frac{{1 + d\left( {x,y} \right)}}{{d\left( {x,y} \right)}} = 1 + \frac{1}{{d\left( {x,y} \right)}} \geqslant 1 + \frac{1}{{d\left( {x,z} \right) + d\left( {z,y} \right)}} = \frac{{1 + d\left( {x,z} \right) + d\left( {z,y} \right)}}{{d\left( {x,z} \right) + d\left( {z,y} \right)}}.$$
So
$$\begin{gathered}
  \frac{{d\left( {x,y} \right)}}{{1 + d\left( {x,y} \right)}} \leqslant \frac{{d\left( {x,z} \right) + d\left( {z,y} \right)}}{{1 + d\left( {x,z} \right) + d\left( {z,y} \right)}} = \frac{{d\left( {x,z} \right)}}
{{1 + d\left( {x,z} \right) + d\left( {z,y} \right)}} + \frac{{d\left( {z,y} \right)}}
{{1 + d\left( {x,z} \right) + d\left( {z,y} \right)}} \hfill \\
   \leqslant \frac{{d\left( {x,z} \right)}}
{{1 + d\left( {x,z} \right)}} + \frac{{d\left( {z,y} \right)}}
{{1 + d\left( {z,y} \right)}} \hfill \\ 
\end{gathered} $$
</s>