Four, Construct $n$ quadratic functions
$$f_{i}(x)=a_{i} x^{2}+2 b_{i} x+c_{i}(i=1,2, \cdots, n) \text {. }$$

Since for each $i$, we have $a_{i}>0, a_{i} c_{i}>b_{i}^{2}$, i.e., $\Delta_{i}=$ $\left(2 b_{i}\right)^{2}-4 a_{i} c_{i} \leqslant 0$, so, $f_{i}(x) \geqslant 0$, that is
$$f_{i}(x)=a_{i} x^{2}+2 b_{i} x+c_{i} \geqslant 0(i=1,2, \cdots, n) .$$

Adding them up, we get $f(x)=f_{1}(x)+f_{2}(x)+\cdots+f_{n}(x) \geqslant 0$,
that is
$$\begin{aligned} f(x)= & \left(a_{1}+a_{2}+\cdots+a_{n}\right) x^{2}+ \\ & 2\left(b_{1}+b_{2}+\cdots+b_{n}\right) x+ \\ & \left(c_{1}+c_{2}+\cdots+c_{n}\right) \geqslant 0 . \end{aligned}$$

Since $a_{1}+a_{2}+\cdots+a_{n}>0$, the parabola opens upwards, and $f(x) \geqslant 0$, so, $\Delta \leqslant 0$. Then we have
$$\begin{array}{l} \Delta=\left[2\left(b_{1}+b_{2}+\cdots+b_{n}\right)\right]^{2}-4\left(a_{1}+a_{2}+\cdots+\right. \\ \left.\quad a_{n}\right)\left(c_{1}+c_{2}+\cdots+c_{n}\right) \leqslant 0 . \\ \text { Hence }\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(c_{1}+c_{2}+\cdots+c_{n}\right) \\ \geqslant\left(b_{1}+b_{2}+\cdots+b_{n}\right)^{2} . \end{array}$$