Let $D\in\mathbb{R}^{n\times n}$ be a real diagonal matrix where $\sum_i D_{ii}<0$. Let also $R\in\mathbb{R}^{n\times n}$ and $L\in\mathbb{R}^{n\times n}$ be real (possibly) non-symmetric (irreducible) matrices with right and left eigenvector $\mathbf{1}$ (associated with eigenvalue 1); i.e. $R\mathbf{1}=\mathbf{1}$ and $\mathbf{1}^T L=\mathbf{1}^T$. How to prove that below equation has a solution $x$ where $|x|>1$? $$ \det\bigg(\underbrace{(L-xI)(R-xI)+(x-1)D}_{\triangleq Q}\bigg)=0. $$
Any hint/help is really appreciated.
Note: In the special scenario when $L=R^T$, one can see that $Q$ is symmetric, and it is not hard to show that $\mathbf{1}^TQ\mathbf{1}<0$ for some $\lambda=1+\epsilon$ with sufficiently small $\epsilon$, therefore $Q$ has at least one negative eigenvalue (eigenvalues are real by symmetricity). Since for $x>>1$ the determinant function $\det(Q)>>0$ (and continuity of eigenvalues), thus there must exists $\lambda^\star>1$ such that the equation holds. One cannot use this approach in the general case, since $Q$ is generally non-symmetric and the numerical range does not coincide with convex hull of the eigenvalues and eigenvalues can be complex (i.e. a negative Reighley quotient doesn't imply existence of a negative eigenvalue).