Let us assume that JD(x,y)>JD(x,z)+JD(z,y)
\(\Rightarrow1-\frac{\left|x\cap y\right|}{\left|x\cup y\right|}>1-\frac{\left|x\cap z\right|}{\left|x\cup z\right|}+1-\frac{\left|z\cap y\right|}{\left|z\cup y\right|}\)
For this to be true \(x\cap y\) has to be non zero i.e it is not an empty set.Since JD(x,y) does not depend on z, thus removing all elements from z would only increase the value of JD(x,z)+JD(z,y) because it decreases the size of the denominator in both the terms therefore not affecting the inequality.Also, we can  remove all the element from z that are in x or y but not in \(x\cap y\)  as this operation would have no effect on inequality because it decreases the value of  \(\frac{\left|x\cap z\right|}{\left|x\cup z\right|}+\frac{\left|z\cap y\right|}{\left|z\cup y\right|}\) . .After the operations we can say that z is a subset of \(x\cup y\)
Now since z is a subset of x  and y
\(RHS=2-\frac{\left|z\right|}{\left|x\right|}-\frac{\left|z\right|}{\left|y\right|}\ \because a\cap b=b\ and\ a\cup b=a\ if\ b\ is\ a\ subset\ of\ a\)