4. Consider the function $$f(x)=\frac{a_{1}}{x}+\frac{a_{2}}{x^{2}}+\cdots+\frac{a_{n}}{x^{n}} .$$ Since $f$ is strictly decreasing from $+\infty$ to 0 on the interval $(0,+\infty)$, there exists exactly one $R>0$ for which $f(R)=1$. This $R$ is also the only positive real root of the given polynomial. Since $\ln x$ is a concave function on $(0,+\infty)$, Jensen's inequality gives us $$\sum_{j=1}^{n} \frac{a_{j}}{A}\left(\ln \frac{A}{R^{j}}\right) \leq \ln \left(\sum_{j=1}^{n} \frac{a_{j}}{A} \cdot \frac{A}{R^{j}}\right)=\ln f(R)=0 .$$ Therefore $\sum_{j=1}^{n} a_{j}(\ln A-j \ln R) \leq 0$, which is equivalent to $A \ln A \leq$ $B \ln R$, i.e., $A^{A} \leq R^{B}$.