Problem 1500

Proof by contradiction. Assume $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}<1$ and $b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}<1$, construct the quadratic function
$$\begin{aligned} f(x) & =(x-1)^{2}-\sum_{k=1}^{n}\left(a_{k} x-b_{k}\right)^{2} \\ & =\left(1-\sum_{k=1}^{n} a_{k}^{2}\right) x^{2}-2\left(1-\sum_{k=1}^{n} a_{k} b_{k}\right) x+1-\sum_{k=1}^{n} b_{k}^{2} \end{aligned}$$

By the assumption $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}<1$, the parabola opens upwards. Given the conditions, its discriminant $\Delta=4\left[\left(1-\sum_{k=1}^{n} a_{k} b_{k}\right)^{2}-\left(1-\sum_{k=1}^{n} a_{k}^{2}\right)\left(1-\sum_{k=1}^{n} b_{k}^{2}\right)\right]<0$, hence $f(x)$ is always greater than 0. However, $f(1)=-\sum_{k=1}^{n}\left(a_{k}-b_{k}\right)^{2} \leqslant 0$, which contradicts $f(x)$ being always greater than 0. Therefore, the assumption is false, and the proposition is proved.