Let $x_1, x_2, . . . , x_n$ be the leaves of a tree $T$. Let $d_{ij}$ be the length of the unique path $P_{ij}$ between $x_i$ and $x_j$ . For any four distinct indices $i$, $j$, $k$ and $l$, show that if $P_{ij}$ and $P_{kl}$ have no vertices in common, then:
$d_{ij} + d_{kl} < d_{ik} + d_{jl} = d_{il} + d_{jk}$
I don't even know how to start this
A hint:
The path $P_{ij}$ contains $\geq2$ vertices $q_r$, and the path $P_{kl}$ contains $\geq2$ vertices $q_r'$. Among all paths connecting a $q_r$ with a $q_r'$ there is a shortest one. It has length $\geq1$ and connects a vertex $q\in P_{ij}$ with a vertex $q'\in P_{kl}$.