Consider the covariance matrix $\mathbf{\Sigma}_{\mathbf{X}+\gamma\mathbf{G}}:=(\mathbf{X}+\gamma\mathbf{G})^{\top}(\mathbf{X}+\gamma\mathbf{G})$ where the rows of $\mathbf{X}, \mathbf{G}\in \mathbb{R}^{n\times m}$ are independently sampled Gaussian random vectors with zero mean (and $\mathbf{X}$ is full-rank, but $\mathbf{G}$ may not be). Now, I want to show that $$ \| \mathbf{\Sigma}_{\mathbf{X}+\gamma\mathbf{G}}^{-1}\mathbf{X}^{\top}\mathbf{y}\|_2 \leq \| \mathbf{\Sigma}_{\mathbf{X}}^{-1}\mathbf{X}^{\top}\mathbf{y}\|_2\;. $$ So basically, the term of the left hand side is bounded by the norm of the OLS solution ($\gamma=0$) on the right hand side.
A relevant previous question can be found here.
Here is what I think about this. Since the eigenvalues of $\mathbf{\Sigma}_{\mathbf{X}+\gamma\mathbf{G}}$ are for sure larger than the respective eigenvalues of $\mathbf{\Sigma}_{\mathbf{X}}$ (because variance is additive, and this holds more generally as well), therefore eigenvalues of $\mathbf{\Sigma}_{\mathbf{X}+\gamma\mathbf{G}}^{-1}$ are lower than the respective eigenvalues of $\mathbf{\Sigma}_{\mathbf{X}}^{-1}$. We can then use the spectral theorem to say that $\| \mathbf{\Sigma}_{\mathbf{X}+\gamma\mathbf{G}}^{-1}\mathbf{z}\|_2 = \sum_i{c_{i,\gamma}(\mathbf{z})^2\lambda_{i, \gamma}^{-2}}$ where $\lambda_{i, \gamma} \geq \lambda_{i, 0}$ are the eigenvalues of $\mathbf{\Sigma}_{\mathbf{X}+\gamma\mathbf{G}}$ and $c_{i, \gamma}(\mathbf{z}):=\mathbf{q_{i, \gamma}}^{\top}\mathbf{z}$ where $\mathbf{q_{i, \gamma}}$ are the orthonormal eigenvectors of $\mathbf{\Sigma}_{\mathbf{X}+\gamma\mathbf{G}}$. What I can't figure out is what is the relationship between $\mathbf{q_{i, \gamma}}$ and $\mathbf{q_{i, 0}}$ (which might help with showing the inequality).
Nevertheless, any other approach that might be able to show that the inequality holds would be helpful!