I have a set of $n$ integers summing up to a known $N$, i.e. $\sum_{i=1}^B N_i = N$. It is very easy to prove that $\sum_{i=1}^B N_i ^ 2 < N^2$, but I want to take a step further. Consider the case in which $\sum_{i=1}^B N_i = (1 + \alpha) N$, for any $\alpha > 0$. For which values of $\alpha$, the above inequality is still true? I tried to proceed by squaring both sides as follows:
$\left(\sum_{i=1}^B N_i\right)^2 = \sum_{i=1}^B N_i ^ 2 + 2 \sum_{j=1}^B \sum_{k < j} N_j N_k = (1 + \alpha)^2N^2$
Rearranging a bit the terms leads to:
$\sum_{i=1}^B N_i ^ 2 = N^2 + \alpha^2N^2 + 2\alpha N^2 - 2 \sum_{j=1}^B \sum_{k < j} N_j N_k$
The inequality holds true if:
$\alpha^2N^2 + 2\alpha N^2 - 2 \sum_{j=1}^B \sum_{k < j} N_j N_k < 0$
which implies:
$\alpha < \frac{\sqrt{N^2 + 2 \sum_{j=1}^B \sum_{k < j} N_j N_k}}{N} - 1$
The result is acceptable, since the square root is always positive and also $\alpha$ is always positive. However, I am wondering if a "better looking" bound can be derived, maybe by substituting previous terms. Also, is there any result related to this problem that I am not aware of? Thank you in advance to anyone will spend time on this.