Problem 1202

$$\begin{array}{l} \text { =, \{50 points\} } \\ \left(a_{1}, b_{1}\right)\left(9\left(a_{2}, b_{2}\right)=\left(a_{1}+a_{2}, b_{1}+b_{2}+a_{1} a_{2}\right)\right. \\ \quad \text { Let }\left(a_{1}, b_{1}\right) \oplus \cdots \oplus\left(a_{k}, b_{k}\right)=\left(a_{1}+a_{2}+\cdots+a_{k}, b_{1}+b_{2}+\cdots+b_{k}\right. \\ \left.+\sum_{1 \leqslant i<j \leqslant k} a_{i} a_{j}\right)(k \geqslant 2 \quad k \in \mathbb{N}) \end{array}$$

Then $\left(a_{1}, b_{1}\right) \oplus \cdots \oplus\left(a_{k}, b_{k}\right) \oplus\left(a_{k+1}, b_{k+1}\right)$
$$\begin{array}{l} =\left(a_{1}+a_{2}+\cdots+a_{k}, b_{1}+b_{2}+\cdots+b_{k}+\sum_{1 \leqslant i<j \leqslant k} a_{i} a_{j}\right) \oplus\left(a_{k+1}, b_{k+1}\right) \\ =\left(a_{1}+a_{2}+\cdots+a_{k+1}, b_{1}+b_{2}+\cdots+b_{k+1}+\sum_{1 \leqslant i<j \leqslant k} a_{i} a_{j}+a_{k+1} \sum_{j=1}^{k} a_{j}\right) \\ =\left(a_{1}+a_{2}+\cdots+a_{k+1}, b_{1}+b_{2}+\cdots+b_{k+1}+\sum_{1 \leqslant i<j \leqslant k+1} a_{i} a_{j}\right) \end{array}$$

Therefore, for $\forall n \in \mathbb{N}, n \geqslant 2$
$$\begin{array}{l} \left(a_{1}, b_{1}\right) \oplus\left(a_{2}, b_{2}\right) \oplus \cdots \oplus\left(a_{n}, b_{n}\right)=\left(a_{1}+a_{2}+\cdots+a_{n}, b_{1}+b_{2}+\cdots+\right. \\ \left.b_{n}+\sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j}\right) \end{array}$$

Therefore, $n\left(\frac{a_{1}+\cdots+a_{n}}{n}, \frac{b_{1}+\cdots+b_{n}}{n}\right)$
$$\begin{array}{l} =\left(a_{1}+a_{2}+\cdots+a_{n}, b_{1}+b_{2}+\cdots+b_{n}+\frac{n-1}{2 n}\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}\right. \\ \end{array}$$

Notice that $2 \sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j}=\left(a_{1}+\cdots+a_{n}\right)^{2}-\sum_{i=1}^{n} a_{i}^{2} \leqslant \frac{n-1}{n}\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}$
Therefore, $\left(a_{1}, b_{1}\right) \oplus\left(a_{2}, b_{2}\right) \oplus \cdots \oplus\left(a_{n}, b_{n}\right) \leqslant n\left(\frac{a_{1}+\cdots+a_{n}}{n}, \frac{b_{1}+\cdots+b_{n}}{n}\right)$