In shortest-path graph, for the unique path $p= ⟨v_0=s,v_1,..,v_k ⟩$ why do we have $d[v_i] \geq d[v_{i-1}] + w(v_{i-1},v_i)$ for $i=1,2,..k$? I was going through the text "Introduction to Algorithms" [2e] by Cormen et.al. where I faced some difficulty in an expression used in the proof. The excerpt from the text is given partially below to explain the context. The expression where I face difficulty is highlighted below.

Lemma 24.17 (Predecessor-subgraph property)
Let $G = (V, E)$ be a weighted, directed graph with weight function $w : E \rightarrow \mathbb{R}$, let $s \in V$ be a source vertex, and assume that $G$ contains no negative-weight cycles that are reachable from $s$. Let us call $\text{Initialize-Single-Source(G, s)}$ and then execute any sequence of relaxation steps on edges of $G$ that produces $d[v] = \delta(s, v)$ for all $v \in V$. Then, the predecessor subgraph $G_\pi$ is a shortest-paths tree rooted at $s$.
Proof.  We must prove that the three properties of shortest-paths ... (the proof starts, I do not show the details).
It remains, therefore, to prove the last property of shortest-paths trees: for each vertex $v \in V_{\pi}$, the unique simple path $p: s\rightsquigarrow v$ in $G_{\pi}$ is a shortest path from $s$ to $v$ in $G$. Let $p = \langle v_0, v_1, ..., v_k \rangle$, where $v_0 = s$ and $v_k = v$. For $i = 1,2,..., k$ , we have both $d[v_i] = \delta(s, v_i)$ and $\require{color}\colorbox{yellow}{$\color{red}d[v_i] \geq d[v_{i-1}] + w(v_{i-1},v_i)$}$, from which we conclude $w(v_{i-1},v_i) \leq \delta(s, v_i) - \delta(s, v_{i-1})$. Summing the weights along path $p$ yields,
$$w(p) = \sum_{i=0}^{k} w(v_{i-1},v_i)$$ $$\leq \sum_{i=0}^{k} (\delta(s, v_i) - \delta(s, v_{i-1}))$$
$$ =...=\delta(s,v_k)$$
Since $\delta(s,v_k)\leq w(p)$ (fact) and $w(p)\leq \delta(s,v_k)$ (shown above), $w(p)=\delta(s,v_k)$

Why there is $\geq$ in the inquation $d[v_i] \geq d[v_{i-1}] + w(v_{i-1},v_i)$ ?

When I tried to find the reason behind it I felt that the authors are assuming that, when the relaxation of the edge $(v_{i-1},v_i)$ happens for the last time, as a result of which $d[v_i]$ gets the actual value of the shortest distance, $\delta(s,v_i)$, during that time :
$$\left. \begin{aligned} \pi[v_i] \leftarrow v_{i-1}\\
d[v_i]\leftarrow d[v_{i-1}]+w(v_{i-1},v_i) \end{aligned} \right \} \tag 1$$
But after the assignments above, $d[v_{i-1}]$ might get changed due to relaxation of edges and $d[v_{i-1}]$ value shall decrease from the one used in the assignment above and hence I think, $d[v_i] \geq d[v_{i-1}] + w(v_{i-1},v_i)$.
But what actually bugs me after this thought is something like this:
When $d[v_i]$ gets the value $\delta(s,v_i)$, from the assignment $d[v_i]\leftarrow d[v_{i-1}]+w(v_{i-1},v_i)$ as shown above, $d[v_{i-1}]$ should have been $\delta(s,v_{i-1})$ at that time, (using the optimal substructure property). Now if it is so, then during subsequent relaxations, $d[v_{i-1}]$ shall retain its value and the inequality in $d[v_i] \geq d[v_{i-1}] + w(v_{i-1},v_i)$ becomes an equality. And I guess this goes in with the ultimate result of the proof which the authors showed...
 A: 
Why is there $\geq$ in the equation $d[v_i] \geq d[v_{i-1}] + w(v_{i-1},v_i)$?
[...]
But after the assignments above, $d[v_{i−1}]$ might get changed due to relaxation of edges and $d[v_{i−1}]$ value shall decrease from the one used in the assignment above and hence I think, $d[v_i] \geq d[v_{i-1}] + w(v_{i-1},v_i)$.

I concur with your analysis, especially since this line of reasoning was also used (with more explanation) in the proof of the previous lemma (see equation 24.12 in Lemma 24.16).

But what actually bugs me after this thought is something like this:
When $d[v_i]$ gets the value $\delta(s,v_i)$, from the assignment $d[v_i] \leftarrow d[v_{i-1}] + w(v_{i-1},v_i)$ as shown above, $d[v_{i−1}]$ should have been $\delta(s,v_{i−1})$ at that time, (using the optimal substructure property). [...]

I don't think we can use the optimal substructure property (Lemma 24.1) here, because we don't know that $p$ is a shortest path from $s$ to $v$.
Nevertheless, you can use your argument to get an alternative proof of Lemma 24.17. Since $d[v_i] = \delta(s,v_i)$ for all $i$, it follows from the triangle inequality (Lemma 24.10) that $d[v_i] \leq d[v_{i-1}] + w(v_{i-1},v_i)$, so we must have equality. By summing over the edges in this path, it is easy to see that $p$ is a shortest path.
I claim that this proof is essentially the same as the proof given in the book. On the one hand, by summing over all edges in the path, we essentially get the computation in the book where it is shown that $w(p) \leq \delta(s,v_k)$. Conversely, we can also deduce from the proof in the book that $d[v_i] = d[v_{i-1}] + w(v_{i-1},v_i)$ for all $i$. This is done in the following way. Using the inequality $w(v_{i-1},v_i) \leq \delta(s,v_i) - \delta(s,v_{i-1})$ and summing over all edges in the path, the authors deduce that $w(p) \leq \delta(s,v_k)$. After that they argue that $w(p) = \delta(s,v_k)$. From this we can deduce that all inequalities in the summation must have been equalities; that is, $w(v_{i-1},v_i) = \delta(s,v_i) - \delta(s,v_{i-1})$, for otherwise we would have $w(p) < \delta(s,v_k)$. Therefore $d[v_i] = d[v_{i-1}] + w(v_{i-1},v_i)$ for all $i$, as promised. Instead of using the triangle inequality to deduce equality right from the start, the authors opted for a more “elementary” proof which only uses the definition of $\delta(s,v_k)$.
The triangle inequality is not really necessary to make the argument work. The proofs are essentially the same; it's just a matter of preference.
