Inverse of sum of matrices has lower norm than the inverse of summand(s) For symmetric $\mathbf{A}, \mathbf{B} \in \mathbb{R}^{m\times m}$ and $\mathbf{c}\in \mathbb{R}^{m\times 1}$, when $\mathbf{A}\succ 0$ and $\mathbf{B}\succeq 0$, I want to show that
\begin{align*}
           \|(\mathbf{A} + \mathbf{B})^{-1}\mathbf{c}\|
        \leq \|\mathbf{A}^{-1}\mathbf{c}\|.
        \end{align*}
I have a feeling it would have something to do with determinants or eigenvalues, but I am not sure how to go about the proof.
 A: Here's a counterexample:
$$
\mathbf A = \left[\begin{matrix}5 & -3\\-3 & 5\end{matrix}\right],\quad 
\mathbf B = \left[\begin{matrix}1 & 0\\0 & 0\end{matrix}\right], \quad 
\mathbf c = \left[\begin{matrix}3\\-4\end{matrix}\right].
$$
We find that
$$
\|(\mathbf{A} + \mathbf B)^{-1}\mathbf c\|^2 \approx \|\mathbf A^{-1}\mathbf c\| + .023
$$
Here's the script I used to generate a "nice" counterexample:
import numpy as np
from numpy.random import randint, rand
from scipy.linalg import norm, inv, det, eig
from sympy import Matrix, latex

for _ in range(100):
    ran = 2
    L = randint(-ran,ran+1,size=[2,2])
    A = L@L.T
    L = randint(-ran,ran+1,size = [2,1])
    B = L@L.T
    try:
        M = inv(A@A) - inv((A+B)@(A+B)) 
    except:
        continue
    if det(M) < -1e-6:
        print(A)
        print(B)
        break
else:
    print('none found')

Once you have a suitable $A$ and $B$, a suitable vector $c$ can be found taking $c$ equal to the eigenvector of $M = A^{-2} - (A+B)^{-2}$ associated with a negative eigenvalue (why?). By rounding the entries of a multiple of this vector $c$, we can get a "nicer" suitable vector $c$.
