Prove this inequality to $\min{(a_{1},a_{2},\cdots,a_{n})}$ Let $a_{1},b_{1},a_{2},b_{2},\cdots,a_{n},b_{n}$ be real numbers such that
$$\min{(a_{1},a_{2},\cdots,a_{n})}\ge 0$$
show that
$$\color{#0a0}{\text{$\max{(a_{1},a_{2},\cdots,a_{n})}\left(\displaystyle\sum_{1\le i,j\le n}\min{(a_{i},a_{j})}b_{i}b_{j}\right)\ge\left(\displaystyle\sum_{1\le i\le n}a_{i}b_{i}\right)^2$}}$$
I have done the proof of the $n=2,3$
$$\min{(a_{1},a_{2},\cdots,a_{n})}\ge 0\Longrightarrow a_{i}\ge 0,i=1,2,\cdots,n$$
Without loss of  generality,Let $a_{1}\ge a_{2}\ge\cdots\ge a_{n}\ge 0$.then we can
$$2a_{1}\sum_{1\le i<j\le n}a_{j}b_{i}b_{j}\ge \left(\sum_{1\le i\le n}a_{i}b_{i}\right)^2$$
but I'm having a problem in proving this inequality.

Assume that $a_1\geq a_i$ for $i>1$, and then
$$
\max{(a_1,...,a_N)}=a_1
$$
$$
\min{(a_1,a_i)}=a_i
$$
The inequality for $(a_1,...,a_N)$ can be transformed into a similar one for $(a_2,...,a_N)$:
$$
a_1 \sum_{i,j=1}^N \min{(a_i, a_j)}b_i b_j \geq \left(\sum_{i=1}^N a_i b_j\right)^2
$$
$$
a_1 \sum_{i,j=2}^N \min{(a_i, a_j)}b_i b_j + 2a_1 b_1 \sum_{i=2}^N a_j b_j+ a_1^2 b_1^2\geq \left(\sum_{i=2}^N a_i b_j\right)^2 + 2a_1b_1\sum_{i=2}^Na_ib_i+a_1^2b_1^2
$$
$$
a_1 \sum_{i,j=2}^N \min{(a_i, a_j)}b_i b_j\geq \left(\sum_{i=2}^N a_i b_j\right)^2
$$
The last inequality is satisfied if $a_1\geq a_i$ and if the initial inequality is satisfied for $(a_2,...,a_N)$. By induction it completes the proof.
