Question 8: Let $n \geq 2$ be a positive integer, and let real numbers $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ satisfy: $\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}=1$. Find the smallest real number $\lambda$ such that $\lambda \cdot \Sigma_{1 \leq i<j \leq n}\left(a_{i} b_{j}-a_{j} b_{i}\right)^{2} \geq \sum_{i=1}^{n}\left(a_{i}-b_{i}\right)^{2}$ always holds.
Eighth question,
Solution: On one hand, take \(a_{1}=a_{2}=\ldots \ldots . .=a_{n-1}=0, a_{n}=1 ; b_{1}=b_{2}=\ldots \ldots=b_{n}=\frac{1}{n}\), then
\[
\begin{array}{l}
\lambda \cdot(n-1) \cdot\left(\frac{1}{n}\right)^{2} \geq(n-1) \cdot\left(\frac{1}{n}\right)^{2}+\left(\frac{n-1}{n}\right)^{2} \\
\Rightarrow \lambda \geq n
\end{array}
\]

On the other hand, we prove: \(n \cdot \sum_{1 \leq i<j \leq n}\left(a_{i} b_{j}-a_{j} b_{i}\right)^{2} \geq \sum_{i=1}^{n}\left(a_{i}-b_{i}\right)^{2}\).
In fact,
\[
\begin{array}{l}
n\left(n \cdot \sum_{1 \leq i<j \leq n}\left(a_{i} b_{i}-a_{j} b_{i}\right)^{2}-\sum_{i=1}^{n}\left(a_{i}-b_{i}\right)^{2}\right) \\
=n^{2}\left[\left(\sum_{i=1}^{n} a_{i}^{2}\right)\left(\sum_{i=1}^{n} b_{i}^{2}\right)-\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{2}\right]-n\left[\left(\sum_{i=1}^{n} a_{i}^{2}\right)+\left(\sum_{i=1}^{n} b_{i}^{2}\right)-2 \sum_{i=1}^{n} a_{i} b_{i}\right] \\
=\left(n \sum_{i=1}^{n} a_{i}^{2}-1\right)\left(n \sum_{i=1}^{n} b_{i}^{2}-1\right)-\left(n \sum_{i=1}^{n} a_{i} b_{i}-1\right)^{2}
\end{array}
\]

Thus, we only need to prove: \(\left(n \sum_{i=1}^{n} a_{i}^{2}-1\right)\left(n \sum_{i=1}^{n} b_{i}^{2}-1\right) \geq\left(n \sum_{i=1}^{n} a_{i} b_{i}-1\right)^{2}\).
Define the quadratic function:
\[
\begin{array}{l}
f(x)=\left(n \sum_{i=1}^{n} a_{i}{ }^{2}-1\right) x^{2}-2\left(n \sum_{i=1}^{n} a_{i} b_{i}-1\right) x+\left(n \sum_{i=1}^{n} b_{i}{ }^{2}-1\right) \\
=n \sum_{i=1}^{n}\left(a_{i} x-b_{i}\right)^{2}-(x-1)^{2} \\
\geq\left(\sum_{i=1}^{n}\left(a_{i} x-b_{i}\right)\right)^{2}-(x-1)^{2}=0
\end{array}
\]

Notice that \(n \sum_{i=1}^{n} a_{i}^{2}-1 \geq\left(\sum_{i=1}^{n} a_{i}\right)^{2}-1=0\), hence its discriminant is no greater than 0, i.e.,
\[
\begin{array}{l}
{\left[2\left(n \sum_{i=1}^{n} a_{i} b_{i}-1\right)\right]^{2}-4\left(n \sum_{i=1}^{n} a_{i}^{2}-1\right)\left(n \sum_{i=1}^{n} b_{i}^{2}-1\right) \leq 0} \\
\Rightarrow\left(n \sum_{i=1}^{n} a_{i}^{2}-1\right)\left(n \sum_{i=1}^{n} b_{i}^{2}-1\right) \geq\left(n \sum_{i=1}^{n} a_{i} b_{i}-1\right)^{2}
\end{array}
\]

In summary, the minimum value of \(\lambda\) is \(n\).
Note: In fact, establish an \(n\)-dimensional coordinate system, set \(A\left(a_{1}, a_{2}, \ldots, a_{n}\right), B\left(b_{1}, b_{2}, \ldots, b_{n}\right)\), then \(\sum_{1 \leq i<j \leq n}\left(a_{i} b_{j}-a_{j} b_{i}\right)^{2}=4 S_{O A B}^{2}, \sum_{i=1}^{n}\left(a_{i}-b_{i}\right)^{2}=|A B|^{2}\), hence \(\frac{\sum_{1 \leq i<i \leq n}\left(a_{i} b_{j}-a_{j} b_{i}\right)^{2}}{\sum_{i=1}^{n}\left(a_{i}-b_{i}\right)^{2}}\) is the square of the length of the perpendicular from 0 to the line \(AB\). This problem is to find the minimum value of the length of the perpendicular from \(O\) to the line \(AB\).
In fact, according to the conditions, \(A\) and \(B\) are both on the "plane" \(\Gamma: x_{1}+x_{2}+\ldots+x_{n}=1\), hence \(\mathrm{d}(0, AB) \geq\)
\[
\mathrm{d}(0, \Gamma)=\frac{\left|\frac{0+0+\ldots+0-1}{n \uparrow}\right|}{\sqrt{\frac{1^{2}+1^{2}+\ldots+1^{2}}{n} \uparrow}}=\frac{1}{\sqrt{n}}, \text{ hence } \lambda \geq n.
\]

Note 2: According to the Lagrange identity, we know:
\[
\begin{array}{l}
\lambda \cdot \sum_{1 \leq i<j \leq n}\left(a_{i} b_{j}-a_{j} b_{i}\right)^{2}-\sum_{i=1}^{n}\left(a_{i}-b_{i}\right)^{2} \\
=\frac{1}{\lambda}\left[\left(\lambda \sum_{i=1}^{n} a_{i}^{2}-1\right)\left(\lambda \sum_{i=1}^{n} b_{i}^{2}-1\right)-\left(\lambda \sum_{i=1}^{n} a_{i} b_{i}-1\right)^{2}\right]
\end{array}
\]