I have a ratio between two determinants: $$r_1 = \frac{\det(\lambda I_d + \sum_{t=1}^{n}x_{t}{x_{t}}^T)}{\det(\lambda I_d)}$$ where $x \in \mathbb{R}^{d}$, $||x||_{2} \leq 1$, $\lambda > 0$, and $I_d \in \mathbb{R}^{d \times d}$ is identity matrix.
Now I add another sum of $n_p$ rank-1 matrices $\sum_{t=1}^{n_p} x_{t}{x_{t}}^T$ (or equivalently write it as ${X_{p}}^{T}X_{p}$, where each row of $X_{p} \in \mathbb{R}^{n_{p} \times d}$ is $x_t$) to both determinants. $$r_2 = \frac{\det(\lambda I_d + \sum_{t=1}^{n}x_{t}{x_{t}}^T + {X_{p}}^{T}X_{p})}{\det(\lambda I_d + {X_{p}}^{T}X_{p})}$$
Now I want to compare the two ratios $r_1$ and $r_2$. Or maybe even better, with some other assumptions I can say $r_1$ is how many times of $r_2$.
Following the generalization of this matrix-determinant lemma, I can rewrite $r_2$ as: $$r_2 = \frac{\det(A + {X_{p}}^{T}X_{p})}{\det(\lambda I_d + {X_{p}}^{T}X_{p})} = \frac{\det(A)\det(I_{n_p}+X_{p} A^{-1} {X_{p}}^{T})}{\det(\lambda I_d)\det(I_{n_p}+ X_{p} (\lambda I_d)^{-1}{X_{p}}^{T})} = r_{1}\frac{\det(I_{n_p}+X_{p} A^{-1} {X_{p}}^{T})}{\det(I_{n_p}+ X_{p} (\lambda I_d)^{-1}{X_{p}}^{T})}$$ where $A=\lambda I_d + \sum_{t=1}^{n}x_{t}{x_{t}}^T$.
So the problem of comparing the two ratios becomes comparing $\frac{\det(I_{n_p}+X_{p} (\lambda I_d + \sum_{t=1}^{n}x_{t}{x_{t}}^T)^{-1} {X_{p}}^{T})}{\det(I_{n_p}+ X_{p} (\lambda I_d)^{-1}{X_{p}}^{T})}$ with $1$.
I don't know what to do from here. Intuitively the numerator should be smaller than denominator, since the term inside the "inverse" in the numerator is "bigger", so $r_2$ should be smaller.