Prove the following inequality $\sum_{i
For positive integer $n \ge 3$, prove the following inequality $$\sum_{i<j<k}\frac{a_ia_ja_k}{(n-2)(n-1)n}\le \bigg(\sum_{i<j}\frac{a_ia_j}{(n-1)n}\bigg)^2+\frac{1}{12}$$ where $a_1+a_2+\cdots +a_n=0$
I noticed that $$(n-2)(n-1)n=6{n \choose 3}$$ and $$(n-1)n=2{n \choose 2}$$
After many arithmetics and research I got: 
$$(n-1)\sqrt[3]{\sum_{i<j<k}\frac{a_ia_ja_k}{n \choose 3}}+\sqrt{\frac{\sum a_i^2}{n}} \le \sum a_i =0$$
Does it help?
Maybe after plugging them into the initial expression, it comes down to apply some famous inequalities that I don't know. Any help is greatly appreciated.
 A: With
$$
 u = \frac{1}{n(n-1)}\sum_{i<j} a_i a_j \, , \quad v =  \frac{1}{n(n-1)(n-2)}\sum_{i<j<k} a_i a_j a_k
$$
the goal to to show that
$$\tag 1
  v \le u^2 + \frac {1}{12} \, .
$$
We can assume that $v > 0$ because otherwise the inequality holds trivially.
The polynomial
$$
 p(x) = (x-a_1)\cdots (x-a_n) = x^n + n(n-1)u x^{n-2} - n(n-1)(n-2)v x^{n-3} + \ldots 
$$
has $n$ real roots. $(n-3)$-fold application of Rolle's theorem shows that the cubic polynomial
$$
 p^{(n-3)}(x) = n(n-1)\cdots 4 \cdot\left( x^3 + 6u x - 6v\right)
$$
has three real roots. It follows that the discriminant $
 \Delta = -4 \left(6u \right)^3 - 27 \left( 6v \right)^2 $ is non-negative, i.e.
$$
\tag 2
v^2 \le -\frac{8}{9} u^3 \, .
$$
It remains to show that $(2)$ implies the desired inequality $(1)$. We see that $u$ must be negative, so that $u = -\sqrt{t}$ for some $t > 0$. Then $(2)$ becomes
$$
 v \le \frac{\sqrt{8}}{3} t^{3/4} 
$$
and in order to get $(1)$ it suffices to show that
$$ 
 \frac{\sqrt{8}}{3} t^{3/4} \le t + \frac{1}{12} \, .
$$
This is an elementary calculation: The difference
$$
f(t) = \frac{\sqrt{8}}{3} t^{3/4} - t - \frac{1}{12}
$$
is maximal at $t^*= 1/4$ with $f(t^*) =0$. This concludes the proof.

One can also see that equality holds exactly if $t=1/4$ and $\Delta = 0$, that is if 
$$
\begin{align}
 u &= \frac{1}{n(n-1)}\sum_{i<j} a_i a_j = -\frac 12 \, ,\\
 v &=  \frac{1}{n(n-1)(n-2)}\sum_{i<j<k} a_i a_j a_k = \frac 13 \, .
\end{align}
$$
