Consider the following system of equations for $R_{i}(\lambda)$ \begin{align} R_1 &= \frac{\lambda}{4}(1 + R_3 + 2 R_2 R_1) \tag{1.1}\\ R_2 &= \lambda \left[q + \left(\frac{1}{2} - q\right)R_2 + q R_1 R_2 + \left(\frac{1}{2} - q \right) R_3 R_2 \right] \tag{1.2} \\ R_3 &= \lambda \left[\frac{1}{2} - q + q R_1 + q R_1 R_3 + \left(\frac{1}{2} - q \right) R_3^2 \right] \tag{1.3} \end{align} where $0 < q < 1/4$. The $R_i(\lambda)$ are generating functions arising in the study of a certain random walk, so they are analytic at least in $|\lambda| \leq 1$.
One can immediately see that there will be a solution to Eqs. $(1)$ with the property that $R_{i}(1) = 1$. Let the set of $R_{i}(\lambda)$ which satisfy Eq. $(1)$ and have $R_{i}(1) = 1$ be called the "basic" solution.
Proposition 1. In the basic solution, there is at least one $i$ such that $R_{i}(\lambda = 1 - \epsilon) > 1$ for some $0 < \epsilon < 1$.
Proof 1. One can see this by reducing Eqs. $(1)$ to a single quartic polynomial in $R_2$ and then solving with Mathematica. The result is a mess, but Proposition 1 is verified directly.
Proof 2. One can compute the derivatives $\text{d} R_i /\text{d}\lambda |_{\lambda = 1 - \epsilon}$ via implicit differentiation and show that they are negative.
Question: I'm interested in showing Proposition 1 for other (bigger) systems like Eqs. $(1)$ so I'm looking for a more powerful proof technique. Is there any way to do this without actually solving the system or computing its derivatives?