If want to prove rigorously that, if there are $n\geq 2$ assignments and the lowest one is dropped, the average will always be greater than that if it was included.
I try to formulate it as such: For $\{a_1,...,a_n\}$, $a_1< a_i\ \forall 1<i\leq n$ and $0\leq a_i\ \forall 1\leq i\leq n$, then $\frac{\sum\limits_{i=1}^n a_i}{n}< \frac{\sum\limits_{i=2}^n a_i}{n-1}$.
Proof attempt: Assume $\frac{\sum\limits_{i=1}^n a_i}{n}\geq \frac{\sum\limits_{i=2}^n a_i}{n-1}$. Then $$(n-1)\sum\limits_{i=1}^n a_i\geq n\sum\limits_{i=2}^n a_i\\ n\sum\limits_{i=2}^n a_i +na_1- \sum\limits_{i=1}^n a_i\geq n\sum\limits_{i=2}^n a_i\\ na_1\geq \sum\limits_{i=1}^n a_i\\ a_1\geq \frac{\sum\limits_{i=1}^n a_i}{n}$$ but I don't believe this implies that $a_1\geq a_i$ so I am stuck here. Thank you for any help.
Hint: try writing the average of $\{a_1,\dots,a_n\}$ as the weighted average of $a_1$ and the average of $\{a_2,\dots,a_n\}$. Then, use the fact that $a_1\le\min(\{a_2,\dots,a_n\})\le\operatorname{avg}(\{a_2,\dots,a_n\})$.