Schedule each entree so that all entrees are completed in the shortest amount of time
The link suggested by @D.W. above put into words exactly what I was looking for, an "exchange lemma" that proves the optimality of the greedy strategy. Here it is:

Suppose we cook in the order $1,2,\ldots, n$. Let $i \in \{1,2,\ldots,n-1\}$ be the smallest integer such that $d_i&lt;d_{i+1}$. We claim that altering the order of the cooking of dishes $i$ and $i+1$, and changing nothing else, results in an equally good or better finishing time. Let us denote $ft(k)$ as the time when dish $k$ is finished under the original ordering (and all dishes are finished at time $\max\limits_{1\leq k\leq n} ft(k)$). We also denote by $ft'(k)$ the time when dish $k$ is finished under the altered ordering.

In general, we have
$$
ft(k) = \sum\limits_{j=1}^k c_j + d_k
$$ since we need to finish cooking dish $k$, which happens after $1,2,\ldots, k-1$, and then finally we dress it.

For the altered ordering we have 
\begin{align*}
ft'(i+1) &amp;= \sum\limits_{j=1}^{i-1} c_j + c_{i+1} + d_{i+1}\\
ft'(i) &amp;= \sum\limits_{j=1}^{i-1} c_j + c_{i+1} +c_i + d_{i}
\end{align*}
(the funny ordering is just to emphasize the order in which the dishes are cooked).

The following three facts are easy to prove:


$ft(k)=ft'(k)$ for all $k \neq i, i+1$
$ft'(i)\leq ft(i+1)$
$ft'(i+1)\leq ft(i+1)$


These three together imply that $\max\{ft'(i),ft'(i+1) \} \leq \max\{ ft(i),ft(i+1)\}$ and hence $\max ft'(k) \leq \max ft(k)$ which proves the greedy strategy.