This inequality maybe is a form of conditional Chebyshev's inequality let $a_{i},b_{i}>0$, show that
$$\sum_{i=1}^{n}a_{i}b_{i}\ge \dfrac{2}{n+\sqrt{\sum_{i=1}^{n}\dfrac{b_{i}}{a_{i}}\sum_{i=1}^{n}\dfrac{a_{i}}{b_{i}}}}\sum_{i=1}^{n}a_{i}\sum_{i=1}^{n}b_{i}\tag{1}$$
I try:since 
$$\sum_{i=1}^{n}\dfrac{b_{i}}{a_{i}}\sum_{i=1}^{n}\dfrac{a_{i}}{b_{i}}\ge n^2$$
We just have to prove the inequality
$$\sum_{i=1}^{n}a_{i}b_{i}\ge\dfrac{1}{n}\sum_{i=1}^{n}a_{i}\sum_{i=1}^{n}b_{i}$$
This looks like a Chebyshev's inequality, but unfortunately the monotonicity of the two sequences is not clear, so it can not be applied directly，so How to prove inequality $(1)$

Proof: We have the following identity:
\begin{align}
&\sum a_i b_i - \frac{2}{n + \sqrt{\sum \tfrac{b_i}{a_i} \sum \tfrac{a_i}{b_i}}} \sum a_i \sum b_i\\
=\ & \sum \frac{b_i}{a_i} 
\left(a_i -  \frac{\frac{a_i}{b_i}\sqrt{\sum \tfrac{b_i}{a_i} \sum \tfrac{a_i}{b_i}} + \sum \frac{a_i}{b_i}}
{\left(n+\sqrt{\sum \tfrac{b_i}{a_i} \sum \tfrac{a_i}{b_i}}\right)\sum \frac{a_i}{b_i}}\sum a_i\right)^2.
\end{align}
We are done.
$\phantom{2}$
Remark: It is not hard to verify the identity above. Let $A = \sum \frac{a_i}{b_i}$, $B = \sum \frac{b_i}{a_i}$, $C = \sum a_i$ and $D = \sum b_i$.
We have
\begin{align}
&\sum \frac{b_i}{a_i}
\left(a_i -  \frac{\frac{a_i}{b_i}\sqrt{\sum \tfrac{b_i}{a_i} \sum \tfrac{a_i}{b_i}} + \sum \frac{a_i}{b_i}}
{\left(n+\sqrt{\sum \tfrac{b_i}{a_i} \sum \tfrac{a_i}{b_i}}\right)\sum \frac{a_i}{b_i}}\sum a_i\right)^2\\
=\ & 
\sum \frac{b_i}{a_i}
\left(a_i - \frac{\frac{a_i}{b_i}\sqrt{AB} + A}{(n+\sqrt{AB})A}C\right)^2\\
=\ &
\sum \frac{b_i}{a_i}
\left(a_i^2 - 2a_i \frac{\frac{a_i}{b_i}\sqrt{AB} + A}{(n+\sqrt{AB})A}C
+ \frac{(\frac{a_i}{b_i}\sqrt{AB} + A)^2}{(n+\sqrt{AB})^2A^2}C^2\right)\\
=\ &
\sum \left(a_ib_i - \frac{2a_i\sqrt{AB} + 2b_iA}{(n+\sqrt{AB})A}C
+ \frac{\frac{a_i}{b_i}AB + 2A\sqrt{AB} + \frac{b_i}{a_i}A^2}{(n+\sqrt{AB})^2A^2}C^2
\right)\\
=\ & \sum a_ib_i - \frac{2C\sqrt{AB} + 2DA}{(n+\sqrt{AB})A}C
+ \frac{A\cdot AB + n\cdot 2A\sqrt{AB} + B\cdot A^2}{(n+\sqrt{AB})^2A^2}C^2\\
=\ &\sum a_ib_i - \frac{2}{n+\sqrt{AB}} CD.
\end{align}
We are done.
