prove the greedy algorithm for Gas Station is valid

Question

I'm trying to prove the solution for the problem Gas Station (LeetCode 134)

Let's define d_i to be the difference g_i - c_i.

Basically it's a greedy solution where you look for the first k s.t.

\sum_{i=k}^n d_i >= 0

Let's denote the following:

A = \sum_{i=1}^k d_i

B = \sum_{i=k+1}^n d_i

Notice that by definition of k, A<0 and B >= 0.

Since we know that there's a unique solution:

A+B >= 0 --> B >= -A

We want to show that B-A >= 0. This will show that we have enough gas for A after finishing the B part.

So B-A >= B+B >= 0

Is my proof correct?

Thanks