Problem: Let $n$ be a positive integer, and let $a_0 \ge a_1 \ge \dots \ge a_n \ge 0$ be integers.
Prove that
\[ \sum_{i=0}^n i\binom{a_i}{2} \le \half \binom{a_0+a_1+\dots+a_n}{2}. \]

Solution: For $n=0$ (which we permit) there is nothing to prove.
Hence to prove by induction on $n$, it would be sufficient to verify
\[ 2n \binom{a_n}{2} \le \binom{a_0 + a_1 + \dots + a_n}{2}
    - \binom{a_0 + a_1 + \dots + a_{n-1}}{2}. \]
Rearranging the terms around, that's equivalent to proving
\begin{align*}
  \iff 2n(a_n^2 - a_n) &\le a_n^2 + a_n \cdot (2(a_0 + \dots + a_{n-1}) - 1) \\
  \iff 0 &\le 2a_n(a_0 + \dots + a_{n-1} - na_n) + a_n(a_n + 2n - 1).
\end{align*}
However, the last line is obvious because $\min(a_0, \dots, a_{n-1}) \ge a_n$,
and $a_n \ge 0$.

\begin{remark*}
  The only equality case is when $a_0 \in \{0,1\}$ and $a_i = 0$ for $i \ge 1$.

  The bound in the problem is extremely loose and pretty much anything will work.
\end{remark*}