Let $\rho_1$ and $\rho_2$ be Hermitian, positive matrices. Is it true that $$\text{Tr}(\sqrt{\sqrt{\rho_1} \rho_2 \sqrt{\rho_1}}) = \text{Tr}(\sqrt{\rho_1 \rho_2}) ~ ?$$ (The expression on the LHS defines the square root of the "fidelity" between $\rho_1$ and $\rho_2$.) Here the square root always means the unique "principal square root".
Here is what I've come up with so far, in favor of the equality: for a matrix with positive eigenvalues, apparently a unique square root, itself having positive eigenvalues, exists. It is true that the product $\rho_1 \rho_2$ has positive eigenvalues (although it's not Hermitian or even diagonalizable in general, I believe), because it has the same eigenvalues (with algebraic multiplicities) as the Hermitian, positive matrix $\sqrt{\rho_1} \rho_2 \sqrt{\rho_1}$. This is because $AB$ and $BA$ always have the same eigenvalues with algebraic multiplicities. Further, the eigenvalues of $\sqrt{A}$ are the same as the square-roots of the eigenvalues of $A$, with algebraic multiplicity included. Finally the trace is always the sum of the eigenvalues, including multiplicity. This seems to show the relation...
This is all correct. $\rho_1 \rho_2$ is similar to $\rho_1^{1/2} \rho_2 \rho_1^{1/2}$, as $$ \rho_1 \rho_2 = \rho_1^{1/2} (\rho_1^{1/2} \rho_2 \rho_1^{1/2}) \rho_1^{-1/2}$$ If $M$ is the positive definite square root of $\rho_1^{1/2} \rho_2 \rho_1^{1/2}$, then $\rho_1^{1/2} M \rho_1^{-1/2}$ is a square root of $\rho_1^{1/2} M^2 \rho_1^{-1/2} = \rho_1 \rho_2$, with the same eigenvalues as $M$. And thus $\text{Tr}(M) = \text{Tr}(\rho_1^{1/2} M \rho_1^{-1/2})$.