### Part (a)
1. Consider the polynomial $p(x) = x^n - a_1 x^{n-1} + a_2 x^{n-2} - \cdots - a_{n-1} x - a_n$.
2. Define the function $f(x) = \sum_{i=1}^n \frac{a_i}{x^i}$. Note that $f(x)$ is a sum of terms of the form $\frac{a_i}{x^i}$, where $a_i \geq 0$ and not all $a_i$ are zero.
3. Since $a_i \geq 0$ and $x > 0$, each term $\frac{a_i}{x^i}$ is positive and $f(x)$ is a strictly decreasing function for $x > 0$.
4. As $x \to 0^+$, $f(x) \to \infty$ because the terms $\frac{a_i}{x^i}$ dominate.
5. As $x \to \infty$, $f(x) \to 0$ because each term $\frac{a_i}{x^i}$ approaches zero.
6. By the Intermediate Value Theorem, since $f(x)$ is continuous and strictly decreasing from $\infty$ to $0$, there exists a unique $R > 0$ such that $f(R) = 1$.

Thus, the polynomial $p(x)$ has precisely one positive real root $R$.

### Part (b)
1. By Jensen's inequality for the convex function $\ln x$, we have:
$$\sum_{i=1}^n \frac{a_i}{A} \ln \left( \frac{A}{R^i} \right) \leq \ln \left( \sum_{i=1}^n \frac{a_i}{A} \cdot \frac{A}{R^i} \right)$$
2. Simplifying the right-hand side, we get:
$$\ln \left( \sum_{i=1}^n \frac{a_i}{R^i} \right) = \ln f(R) = \ln 1 = 0$$
3. Therefore, we have:
$$\sum_{i=1}^n \frac{a_i}{A} \ln \left( \frac{A}{R^i} \right) \leq 0$$
4. Expanding the left-hand side, we get:
$$\sum_{i=1}^n \frac{a_i}{A} \left( \ln A - i \ln R \right) = \ln A \sum_{i=1}^n \frac{a_i}{A} - \ln R \sum_{i=1}^n \frac{ia_i}{A}$$
5. Since $\sum_{i=1}^n \frac{a_i}{A} = 1$, this simplifies to:
$$\ln A - \frac{\ln R}{A} \sum_{i=1}^n ia_i \leq 0$$
6. Let $B = \sum_{i=1}^n ia_i$. Then:
$$\ln A \leq \frac{B \ln R}{A}$$
7. Multiplying both sides by $A$, we get:
$$A \ln A \leq B \ln R$$
8. Exponentiating both sides, we obtain:
$$A^A \leq R^B$$

The final answer is $\boxed{ A^A \leq R^B }$