Was there a "nice" way of doing this? The really weird equality case $a_1 = a_2 = a_3 = \dots = a_{n - 1} = \frac{1}{n - 1}, a_n = n - 1$ and the non-obvious value for $c$ (namely $(n^2 - n)/(n^2 - n + 1)$) lead me to believe there may not have been. I have like 70 percent of a really ugly convexity + Lagrange Multipliers + bashing solution but if someone on the ELMO committee knows of a better (or the intended) one I would love to see it.

Here is elementary proof for this nice inequality. If we denote $d=\frac{n}{c}$. We will prove, using induction, that for all $d\ge n+\frac{1}{n-1}$, equivalent inequality: $\sum_{i=1}^{n} \frac{1}{d+a_i^2}\le \frac{n}{d+1}$ holds. Base $n=2$: We need to prove that for $a+b=2$ and $d\ge 3$, inequality: $\frac{1}{d+a^2}+\frac{1}{d+b^2}\le \frac{2}{d+1}$ holds. $\Leftrightarrow 2a^2b^2+d(a^2+b^2-2)-a^2-b^2\ge 0$. Since LHS is increasing in $d$, it's enough to prove inequality for $d=3$, but this is equivalent to: $2(ab-1)^2\ge 0$ Now suppose statement holds for $n$ and we will prove for $n+1$, i.e.: $\sum_{i=1}^{n+1} \frac{1}{d+a_i^2}\le \frac{n+1}{d+1}$ for $d\ge n+1+\frac{1}{n}$ WLOG we can suppose $a_1\le ...\le a_{n+1}$. Then $a_1+...+a_n=tn$, $t\le 1$ Now, let's define $b_i=\frac{a_i}{t}$, for all $i=1,...,n$, so $b_1+...+b_n=n$ and let's define $d'=\frac{d}{t^2}>n+\frac{1}{n-1}$ From assumption, it follows: $\sum_{i=1}^{n} \frac{1}{d'+b_i^2}\le \frac{n}{d'+1}$. Therefore: $\sum_{i=1}^{n} \frac{1}{d+a_i^2}= \frac{1}{t^2}\sum_{i=1}^{n} \frac{1}{d'+b_i^2}\le \frac{n}{d+t^2}$. Finally, it remains to prove: $\frac{n}{d+t^2}+\frac{1}{d+a_{n+1}^2}\le \frac{n+1}{d+1}$. What is, using $a_{n+1}=n+1-nt$ equivalent to: $n(n+1)(t-1)^2(d+nt^2-2t-n-1)\ge 0$. But since $d\ge n+1+\frac{1}{n}$,: $LHS \ge (n+1)(t-1)^2(nt-1)^2 \ge 0$. Hence, inequality is proved. Equality holds when $a_1=...=a_n=1$ and when $a_1=...=a_{n-1}=\frac{1}{n-1}, \; a_n=n-1$. $\blacksquare$

write the given condition as $\sum_{i=1}^{n} a_{i} - 1 = 0$ and write the ineqn as : $$\sum_{i=1}^{n} \frac{1}{n+c(a_{i}^{2})} - \frac{1}{n+c} \le {0}$$then we have $\sum_{i=1}^{n} \frac{c-c({a_{i}}^{2})}{(n+c)(n+c({{a_{i}}^{2}))}} \le {0}$ or we have $$\sum_{i=1}^{n} \frac{1-(a_{i})^{2}}{n+c({a_{i}}^{2})} \le {0}$$now W.L.O.G ${a_{1}}\ge{a_{2}}\ge..........\ge{a_{n}}$ also our ineqn can be written as $$\sum_{i=1}^{n} (1-a_{i})(\frac{1+a_{i}}{n+c({a_{i}}^{2})} \le {0}$$note for this to happen by chebyshev's : ${1-a_{i}}\le{1-a_{2}}\le........\le{1-a_{n}}$ (sequence 1) and ${\frac{1+a_{1}}{n+c({a_{1}}^{2})}}\ge{\frac{1+a_{2}}{n+c({a_{2}}^{2})}}\ge.........\ge{\frac{1+a_{n}}{n+c({a_{n}}^{2})}}$ (sequence 2) which would give us our ineqn : so sequence 1 ordering is true by WLOG we just need to have a 'c' which makes true .. so now we have : ${\frac{1+a_{1}}{n+c({a_{1}}^{2})}}\ge{\frac{1+a_{2}}{n+c({a_{2}}^{2})}}$ our we now arrange these terms involved in the above frac: we get : ${n}\ge{c(a_{1}+a_{2}+a_{1}a_{2})}$ similarly ${n}\ge{c(a_{2}+a_{3}+a_{2}a_{3})}$ and at last conutuniung till $n-1$ we have ${n}\ge{c(a_{n-1}+a_{n}+a_{n-1}a_{n})}$ now sum all these ineqn : we have $${n(n-1)}\ge(c)({n+(a_{2}+a_{3}....... a_{n}) + (a_{1}a_{2} ..... a_{n-1}a_{n}))}$$now by abel sum : we would have $${\frac{n^{2}-n}{n^{2}-n+1}}\ge{c}$$or we have $$c_{max}=\frac{n^{2}-n}{n^{2}-n+1}$$.. Equality holdes when

$a_1=...=a_n=1$ and $a_1=...=a_{n-1}=\frac{1}{n-1}, \; a_n=n-1$ Abel