Suppose I have symmetric matrices $A \in \mathbb{R}^{n \times n}$ and $B \in \mathbb{R}^{n \times n}$ which are both positive definite. I am wondering if I one can bound ${\rm tr}\left(A - B \right)$ in the following way:
\begin{align*} {\rm tr}\left(A - B \right) & = {\rm tr}\left[ (A^{1/2} - B^{1/2})(A^{1/2} + B^{1/2}) \right] \\ & \leq {\rm tr}\left[ (A^{1/2} - B^{1/2})\right] f(A^{1/2} + B^{1/2}), \end{align*}
for some function $f$, e.g., spectral norm? Does such an inequality exist?
In full generality, I cannot say anything about the sign of the eigenvalues of $A^{1/2} - B^{1/2}$, so as far as I can tell, many of the standard inequalities do not apply.
Any insight would be very helpful.
Edit: An example showing that such an inequality will fail in certain cases was suggested by Darij Grinberg in the comments below.
One way to get your inequality: let $e_i$ be an orthonormal eigenbasis of $A^{1/2} + B^{1/2}$. We have $$ {\rm tr}\left[ (A^{1/2} - B^{1/2})(A^{1/2} + B^{1/2}) \right] = \sum_{i=1}^n e_i^T(A^{1/2} - B^{1/2})(A^{1/2} + B^{1/2})e_i =\\ \sum_{i=1}^n \lambda_i e_i^T (A^{1/2} - B^{1/2}) e_i \leq \sum_{i=1}^n \|A^{1/2} + B^{1/2}\| \, e_i^T (A^{1/2} - B^{1/2}) e_i =\\ \|A^{1/2} + B^{1/2}\| \operatorname{tr}(A^{1/2} - B^{1/2}) $$
Another inequality that may interest you: because $A,B \mapsto \operatorname{tr}(AB)$ forms an inner product over the symmetric matrices, we may use the Cauchy-Schwarz inequality to conclude that $$ {\rm tr}\left[ (A^{1/2} - B^{1/2})(A^{1/2} + B^{1/2}) \right] \leq \sqrt{{\rm tr}[(A^{1/2} - B^{1/2})^2]} \sqrt{{\rm tr}[(A^{1/2} + B^{1/2})^2]} $$