Let $a_{1}, \cdots, a_{n}, b_{1}, \cdots, b_{n}$ be positive real numbers. Prove that:
$$ \left( \sum\limits_{i\neq j}a_{i}b_{j} \right)^2 \geq \left( \sum\limits_{i\neq j}a_{i}a_{j} \right) \left( \sum_{i\neq j}b_{i}b_{j} \right) $$
One solution is:
Let us denote:
$p = \sum_{i=1}^{n}a_{i}, q = \sum_{i=1}^{n}b_{i}, k = \sum_{i=1}^{n}a_{i}^2, l = \sum_{i=1}^{n}b_{i}^2, m = \sum_{i=1}^{n}a_{i}b_{i}$
Then:
$\sum_{i \neq j}a_{i}b_{j} = pq - m, \sum_{i \neq j}a_{i}a_{j} = p^2 - k, \sum_{i \neq j}b_{i}b_{j} = q^2 - l$
So, the required inequality is equivalent to:
$(pq-m)^2 \geq (p^2 - k)(q^2-l) \iff lp^2-2qm.p + m^2 + q^2k - kl \geq 0$
If we prove that its discriminant is less than or equal to 0, we are done. That condition can be written as:
$q^2m^2 -l(m^2+q^2k -kl) \leq 0 \iff (lk-m^2)(q^2-l) \geq 0$
The last inequality is true because $q^2-l = \sum_{i \neq j}b_{i}b_{j} \geq 0 (b_{i}$ are positive), and $lk-m^2 \geq 0 $ (Cauchy-Schwarz inequality) The equality holds if and only if $lk - m^2 = 0$
Now, I'm looking for other solutions to prove it, please comment on