show inequality $\frac{1}{128}\ge \sum_{1\le i\neq j\leq n}a_i^5a_j^3$ Let $a_1,...a_n$ be positive reals with sum 1. Show that 
$$\frac{1}{128}\ge \sum_{1\le i\neq j\leq n}a_i^5a_j^3$$
It seems hard to show it .
 A: We'll prove the statement for nonnegative reals. Firstly, note that for any $a_i, a_j$ we have $(a_i-a_j)^2\ge 0\implies a_ia_j\le \frac{1}{4}(a_i+a_j)^2\le \frac{1}{4}$. Hence, the sum at question is at most $\frac{1}{16}\sum_{i\neq j}a_i^3a_j=\frac{1}{16}\sum_i a_i^3(1-a_i)$, so it suffices to check that $\sum_i a_i^3(1-a_i)\le \frac{1}{8}$ when $a_i$ sum to $1$. This we may do by induction on $n$. For the base case $n=2$, this is just $8a_1a_2(a_1^2+a_2^2)\le 1=(a_1+a_2)^4\Leftrightarrow (a_1-a_2)^4\ge 0$. 
Now suppose the  inequality holds for $n=k$. Suppose $a_1, \ldots, a_{k+1}$ have sum one, and WLOG suppose $a_1\le a_2\le \cdots \le a_{k+1}$ so that $a_1+a_2\le \frac{2}{n}\le \frac{3}{4}$. Then we claim that the sum in question increased under the transformation $(a_1, a_2)\to (0, a_1+a_2)$. Indeed, the difference of the sum after this transformation and before it is:
$$(a_1+a_2)^3-a_1^3-a_2^3-(a_1+a_2)^4+a_1^4+a_2^4=a_1a_2(3a_1+3a_2-4a_1^2-6a_1a_2-4a_2^2)\ge a_1a_2(4(a_1+a_2)^2-4a_1^2-6a_1a_2-4a_2^2)\ge 0$$
So, it suffices to check that $\sum_{i=1}^kb_i^3(1-b_i)\le \frac{1}{8}$, where $b_1=a_1+a_2, b_i=a_{i+1}, 2\le i\le k$. But this is covered by the inductive hypothesis (after reindexing the $b_i$), so the induction is complete and we're done.
Equality holds when two of the variables are $\frac{1}{2}$ and the rest are $0$.
