Lemma 1. If the triangle inequality holds in a graph with positive weights, then d(a, b) is at most the weight of any path from a to b.
Proof. We prove the lemma by induction. Let the inductive hypothesis P (n) be the statement “If n is the length of a path p from a to b, for any vertices a and b, then d(a, b) is at most the weight of p.”
This holds for the base case of n = 1, since the minimum cost of traveling from a to b cannot be strictly greater than the weight of the direct path from a to b (otherwise, we could simply take the direct path for a smaller cost).
In the above proof I know the base case is true because if length of path $p$ is $1$, then $d(a, b)= \text{weight of } p$.
However I can't understand their reasoning which states: $\text{minimum cost of travelling from a to b } \le d(a,b)$
EDIT: Graph is complete graph and $d(a,b)$ is known for each pair of vertices, where $d(a,b)$ is the weight of edge $a-b$.
This follows from repeated application of triangle inequality (without explicitly using induction). Indeed, if you have any path $p$ from $a$ to $b$ consisting of $n$ edges $p_i$ connecting vertices $c_{i-1}$ to $c_i$ (hence $c_0=a, c_n=b$), and if $w$ denotes the weight function, then you have
$$d(a,b) \le \sum_i d(c_{i-1},c_i) = \sum_i w(p_i) = w(p).$$