Given below is an excerpt from the text A First Course in Graph Theory by Chartrand, Zhang.
Here $L(W)$ means the length of the walk $W$.Graph $G$ has $n$ vertices. $h(G)$ is the length of the Hamiltonian walk in $G$.
Let $W$ be a Hamiltonian walk in $G$ with $L(W)=h(G)$. Suppose that $W=(x_1,x_2,...,x_N,x_1)$, where $N\geq n$. Define $v_1=x_1$ and $v_2=x_2$. For $3\leq i \leq n$, define $v_i$ to be $x_{j_i}$, where $j_i$ is the smallest positive integer such that $x_{j_i} \notin \{v_1,v_2,...,v_{i-1}\}$. Then $s: v_1,v_2,...,v_n,v_{n+1}=v_1$ is cyclic ordering of $V(G)$. For each $i$ with $1\leq i\leq n$, let $Q_i$ be the $v_i-v_{i+1}$ subwalk of $W$ and so $d(v_i,v_i+1) \leq L(Q_i)$. So,
$$\sum_{i=1}^{n} L(Q_i)\leq L(W) \tag 1$$
I cannot understand why in $(1)$ we are getting a possibility of $\lt$ ?
The possibility doesn't exist, since for any cyclic ordering of the vertices, if we take the concatenation of geodesics between consecutive vertices we get a closed walk that passes through all the vertices, and therefore its length is less than $L(W)$.
I think in this part of the book the author simply wants to show that in order to find a hamiltonian walk for the graph it is sufficient to just consider all cyclic orderings of the vertices, a hamiltonian walk can be obtained by considering the concatenations of geodesics between consecutive vertices in the ordering.