A first step could be the following: A corollary to Weyl's eigenvalue inequalities (see, e.g., Horn & Johnson, second edition, Corollary 4.3.15), is that the largest eigenvalue of a sum of matrices is as least as large as the sum of the largest eigenvalue of one of the matrices and the smallest eigenvalue of the other matrix, i.e.,
$$ \lambda_M(R_1+R_2)\ge\lambda_M(R_1)+\lambda_1(R_2) $$
Since the matrices are covariance matrices, they are positive semidefinite, and hence $\lambda_1(R_2)\ge0$.
Note that you can easily omit the identity matrix by just adding it to the matrix $R_2$. Note further that, since all matrices involved are symmetric,
$$ R_1(R_1+R_2)^{-2}R_1 = R_1(R_1+R_2)^{-1}(R_1+R_2)^{-1}R_1 = R_1(R_1+R_2)^{-1}\left(R_1(R_1+R_2)^{-1}\right)^T. $$
Hence, for the moment it might suffice to determine the largest eigenvalue of $R_1(R_1+R_2)^{-1}$ and then use the Corollary to Gelfand's formula to bound
$$ \lambda_M\left(R_1(R_1+R_2)^{-1}\left(R_1(R_1+R_2)^{-1}\right)^T\right) \le \lambda_M(R_1(R_1+R_2)^{-1})^2. $$
Best Answer
To repeat the Wikipedia article on positive-definite matrix:
Suppose the complex matrix $M$ is positive definite. The diagonal entries $m_{ii}$ are real and non-negative. As a consequence the trace, $tr(M) ≥ 0$. Furthermore, since every principal sub matrix (in particular, $2\times 2$) is positive definite, $${\displaystyle |m_{ij}|\leq {\sqrt {m_{ii}m_{jj}}}\leq {\frac {m_{ii}+m_{jj}}{2}}}$$ and thus $${\displaystyle \max |m_{ij}|\leq \max |m_{ii}|}$$