Question
$$\text{Let } b_{i}\wedge a_{i}>0 \text{ where } i\in{1,2,3,\ldots,n} \nonumber , \sum_{i=1}^{n}(b_{i}) = \lambda \text{ then Prove that} \nonumber \frac{\lambda-(b_{1}+b_{2})}{(b_{1}+b_{2})}(a_{1}+a_{2})+\frac{\lambda-(b_{1}+b_{3})}{(b_{1}+b_{3})}(a_{1}+a_{3})+\ldots+\frac{\lambda-(b_{2}+b_{3})}{(b_{2}+b_{3})}(a_{2}+a_{3})+\ldots \frac{\lambda-(b_{n-1}+b_{n})}{(b_{n-1}+b_{n})}(a_{n-1}+a_{n}) \nonumber \ge \sqrt{\frac{n(n-1)(n-2)^{2}}{4}\sum_{1\le i<j\le n}(a_{i}a_{j})} \nonumber$$
My work so far
$$(x_1y_1 + x_2y_2 + \ldots + x_ny_n)^2 \leq (x_1^2 + x_2^2 + \ldots + x_n^2)(y_1^2 + y_2^2 + \ldots + y_n^2)$$
$x_i = \frac{\lambda - (b_i + b_j)}{(b_i + b_j)}$ $y_i = a_i$ for some $j \neq i$
$$\sum_{i=1}^{n} \sum_{j \neq i} \left(\frac{\lambda - (b_i + b_j)}{(b_i + b_j)}\right)^2 (a_i)^2 \ge \left(\sum_{i=1}^{n} \sum_{j \neq i} \frac{\lambda - (b_i + b_j)}{(b_i + b_j)}(a_i)\right)^2$$
$$\sum_{i=1}^{n} \sum_{j \neq i} \frac{(\lambda - (b_i + b_j))^2}{(b_i + b_j)^2}(a_i)^2 \ge \left(\sum_{i=1}^{n} \sum_{j \neq i} \frac{\lambda - (b_i + b_j)}{(b_i + b_j)}(a_i)\right)^2$$
$$\sum_{i=1}^{n} \sum_{j \neq i} \frac{\lambda^2 - 2\lambda(b_i + b_j) + (b_i + b_j)^2}{(b_i + b_j)^2}(a_i)^2 \ge \left(\sum_{i=1}^{n} \sum_{j \neq i} \frac{\lambda - (b_i + b_j)}{(b_i + b_j)}(a_i)\right)^2$$