let there be an element $a$ that's in exactly $k$ sets. this element will add to the first sum $k$, to the second sum ${k}\choose{3}$ and to the last sum ${k}\choose{2}$. if we prove te inequality by doing it one element at a time, then at the end it'll be true (since with start with empty set, which gives $0 \ge 0$). so we want to prove that: $\frac{k}{n} + \frac{{{k}\choose{3}}}{{{n}\choose{3}}} \ge \frac{2{{k}\choose{2}}}{{{n}\choose{2}}}$ $\frac{k}{n} + \frac{k(k-1)(k-2)}{n(n-1)(n-2)} \ge \frac{2k(k-1)}{n(n-1)}$ $1 + \frac{(k-1)(k-2)}{(n-1)(n-2)} \ge \frac{2(k-1)}{n-1}$ $(n-1)(n-2) + (k-1)(k-2) \ge 2(k-1)(n-2)$ let $n = 2 + s$ with $s \ge 1$ and $k = 1 + t$ with $t \ge 0$ $(s+1)s + t(t-1) \ge 2ts$ $(t-s)^2 \ge t-s$ which is true, since $t-s$ is an integer

Thats exactly the same solution I found I just proved the last step directly without the substitution Yimin