$\left(\sum_i a_i\right)^2\ge (n-1)\sum_i a_i^2 + b\implies 2a_i a_j \ge b/(n-1) \quad (\forall i\ne j)$ 
Let $a_1, \dots, a_n(n\ge 2), b$ be real numbers, and
$$
\left(\sum_{i=1}^n a_i\right)^2\ge (n-1)\sum_{i=1}^n a_i^2 + b,
$$
Then we have
$$
2a_i a_j \ge b/(n-1) \quad (\forall i\ne j)
$$

$n=2$, clear.
Suppose that for some $n>2$, we have
$$
\left(\sum_{i=1}^n a_i\right)^2\ge (n-1)\sum_{i=1}^n a_i^2 + b\implies 2a_i a_j \ge b/(n-1) \quad (\forall i\ne j).
$$
I want to show that
$$
\left(\sum_{i=1}^{n+1} a_i\right)^2\ge n\sum_{i=1}^{n+1} a_i^2 + b\implies 2a_i a_j \ge b/n \quad (\forall i\ne j),
$$
but it is a little diffcult for me to show
$
\left(\sum_{i=1}^{n+1} a_i\right)^2\ge n\sum_{i=1}^{n+1} a_i^2 + b\implies \left(\sum_{i=1}^{n} a_i\right)^2\ge (n-1)\sum_{i=1}^{n} a_i^2 + b
$
first.
Could you please give me a better solution? Thanks!
 A: Let $n>2$, $b>2(n-1)a_ia_j,$ and $\sum\limits_{k\neq i,k\neq j}a_k=t$
Thus, $$\left(\sum_{k=1}^na_k\right)^2\geq(n-1)\sum_{k=1}^na_k^2+b>(n-1)\sum_{k=1}^na_k^2+2(n-1)a_ia_j,$$ which is a contradiction because we'll prove now that
$$(n-1)\sum_{k=1}^na_k^2+2(n-1)a_ia_j\geq\left(\sum_{k=1}^na_k\right)^2.$$
Indeed, by C-S $$(n-1)\sum_{k=1}^na_k^2+2(n-1)a_ia_j=(n-1)(a_i+a_j)^2+\frac{n-1}{n-2}\cdot(n-2)\sum\limits_{k\neq i,k\neq j}a_k^2\geq$$
$$\geq(n-1)(a_i+a_j)^2+\frac{n-1}{n-2}\cdot t^2.$$
Thus, it's enough to prove that:
$$(n-1)(a_i+a_j)^2+\frac{n-1}{n-2}\cdot t^2\geq(a_i+a_j+t)^2$$ or
$$(n-2)(a_i+a_j)^2+\frac{1}{n-2}\cdot t^2\geq2(a_i+a_j)t,$$ which is true by AM-GM.
Id est, it's enough to solve our problem for $n=2$, which you made.
A: We can convert @MichaelRozenberg's proof into a direct argument: If $i\ne j$ then
$$
\sum a_k^2+2a_ia_j=(a_i+a_j)^2+\sum_{k\not\in\{i,j\}}a_k^2\stackrel{(1)}\ge(a_i+a_j)^2+\frac{\left[(\sum a_k)-a_i-a_j\right]^2}{n-2}\stackrel{(2)}\ge\frac{(\sum a_k)^2}{n-1}\tag a
$$
where (1) is Cauchy-Schwarz and (2) is the inequality $x^2+\dfrac{y^2}\alpha\ge\dfrac{(x+y)^2}{\alpha+1}$ which follows from $\alpha x^2+\dfrac{y^2}\alpha\ge 2xy$ when $\alpha>0$. So if $b$ is such that
$$
\dfrac{(\sum a_k)^2}{n-1}\ge\sum a_k^2 + \frac b{n-1}\tag b
$$
then (a) and (b) imply
$2a_ia_j\ge\frac b{n-1}$
for every $i\ne j$.
