1-norm and symmetry Define the fidelity function for positive operators by $F(\rho, \sigma) = \lVert \sqrt{\rho}\sqrt{\sigma}\rVert_1$. Here, $\lVert\cdot\rVert_1$ is the Schatten 1-norm and defined as $\lVert A\rVert_1 = \operatorname{Tr}(\sqrt{A^{\dagger}A})$.
I'm having some trouble showing that $F$ is symmetric in its arguments. Physics textbooks do it through the idea of purifications but I thought there should be a mathematical argument. 

How does one see that $F(A,B) = F(B,A)$ given that $F(\rho, \sigma) = \operatorname{Tr}(\sqrt{\sqrt{\rho}\sigma\sqrt{\rho}})$?

My original idea to prove it was to try and use cyclicity of trace and assume that $\sqrt{AB} = \sqrt{A}\sqrt{B}$, but it seems this is invalid even for positive operators? Can someone also comment on this seemingly simple statement being false?

We know that $F( \rho, \sigma) = \text{Tr}( \sqrt{ \sqrt{\rho} \sigma \sqrt{\rho}})$. Now consider
\begin{align*}
F( \sigma, \rho) &= \lVert \sqrt{\sigma} \sqrt{\rho} \rVert_1 = \text{Tr} \big( \sqrt{(\sqrt{\sigma} \sqrt{\rho})^\dagger \sqrt{\sigma} \sqrt{\rho}} \big) \\
&= \text{Tr}\big( \sqrt{ \sqrt{\rho} \sigma \sqrt{\rho}} \big) = F( \rho, \sigma). 
\end{align*}
