Let $A$, $B$ and $C$ be symmetric, positive semi-definite matrices. Is it true that $$ \|(A + C)^{1/2} - (B + C)^{1/2}\| \leq \|A^{1/2} - B^{1/2}\|,$$ in either the 2 or Frobenius norm?
It is clearly true when $A, B$ and $C$ commute, but the general case is less clear to me. In fact, even the particular case $B = 0$ does not seem obvious.
Without loss of generality, it is clear that we can assume that $C$ is diagonal. We show that it is sufficient to prove to prove the inequality for the matrix with zeros everywhere except on any position $k$ on the diagonal, $$ (C_k)_{ij} = \begin{cases} 1 & \text{if } i=j=k\\ 0 & \text{otherwise} \end{cases} $$ Clearly, if the inequality is true for one $C_k$, it is true for any $C_k$, by flipping the axes, and also for $C = \alpha C_k$, for any $\alpha \geq 0$, because \begin{align} \|(A + \alpha \, C_k)^{1/2} - (B + \alpha C_k)^{1/2}\| &= \sqrt{\alpha} \|(A/\alpha + C_k)^{1/2} - (B/\alpha + C_k)^{1/2}\| \\ &\leq \sqrt{\alpha} \|(A/\alpha)^{1/2} - (B/\alpha)^{1/2}\| = \sqrt{\alpha} \|A^{1/2} - B^{1/2}\| \end{align} Now, a general diagonal $C$ can be decomposed as $C = \sum_{k=1}^{n} \alpha_k C_k$. Applying the previous inequality (specialized for a matrix $C$ with only one nonzero diagonal element) repeatedly, we can remove the diagonal elements one by one \begin{align} &\|(A + \sum_{k=1}^{n}\alpha_k \, C_k)^{1/2} - (B + \sum_{k=1}^{n}\alpha_k \, C_k)^{1/2}\| \\ &\qquad = \|((A + \sum_{k=1}^{n-1}\alpha_k \, C_k) + \alpha_n C_n)^{1/2} - ((B + \sum_{k=1}^{n-1}\alpha_k \, C_k) + \alpha_n C_n)^{1/2}\| \\ &\qquad \leq \|(A + \sum_{k=1}^{n-1}\alpha_k \, C_k)^{1/2} - (B + \sum_{k=1}^{n-1}\alpha_k \, C_k)^{1/2}\| \\ &\qquad \leq \|(A + \sum_{k=1}^{n-2}\alpha_k \, C_k)^{1/2} - (B + \sum_{k=1}^{n-2}\alpha_k \, C_k)^{1/2}\| \\ &\qquad \leq \dots \leq \sqrt{\alpha} \|A^{1/2} - B^{1/2}\|. \end{align}
Here are three ways of proving the inequality in 1 dimension, which I tried to generalize to the multidimensional case without success. Let us write $a$, $b$, $c$ instead of $A$, $B$, $C$, to emphasize that we are working in one dimension, and let us assume without loss of generality that $a \leq b$.
Let us write: $$ f(c) = \sqrt{b + c} - \sqrt{a + c} $$ We calculate that the derivative of $f$ is given by $$ f'(c) = \frac{1}{2} \left( \frac{1}{\sqrt{b + c}} - \frac{1}{\sqrt{a + c}} \right) \leq 0, $$ and so $f(c) = f(0) + \int_{0}^{c} f'(x) \, d x \leq f(0)$.
We have, by the fundamental theorem of calculus and a change of variable \begin{align} \sqrt{b + c} - \sqrt{a + c} &= \int_{a + c}^{b + c} \frac{1}{2 \sqrt{x}} \, d x = \int_{a}^{b} \frac{1}{2 \sqrt{x + c}} \, d x \\ &\leq \int_{a}^{b} \frac{1}{2 \sqrt{x}} \, d x = \sqrt{b} - \sqrt{a}. \end{align}
Squaring the two sides of the inequality, we obtain $$ a + c - 2 \sqrt{a+ c} \, \sqrt{b + c} + b + c \leq a + b - 2 \sqrt{a} \sqrt{b}. $$ Simplifying and rearranging, $$ c + \sqrt{a} \sqrt{b} \leq \sqrt{a+ c} \, \sqrt{b + c} . $$ Squaring again $$ \require{cancel} \cancel{c^2 + a b} + 2 c \sqrt{a b} \leq \cancel{c^2 + ab} + ac + bc, $$ leading to $$ a + b - 2 \sqrt{ab} = (\sqrt{b} - \sqrt{a})^2 \geq 0$$.
Numerical experiments suggest that the inequality is true in both the 2 and the Frobenius norm. (One realization of) the following code prints 0.9998775.
import numpy as np
import scipy.linalg as la
n, d, ratios = 100000, 3, []
for i in range(n):
A = np.random.randn(d, d)
B = np.random.randn(d, d)
C = .1*np.random.randn(d, d)
A, B, C = A.dot(A.T), B.dot(B.T), C.dot(C.T)
lhs = la.norm(la.sqrtm(A + C) - la.sqrtm(B + C), ord='fro')
rhs = la.norm(la.sqrtm(A) - la.sqrtm(B), ord='fro')
ratios.append(lhs/rhs)
print(np.max(ratios))
