= Solution
Use the unsquared <quantum fidelity> convention in the question. For any <matrix> $X$, its <trace norm> is $\|X\|_1=\operatorname{Tr}\sqrt{X^\dagger X}=\operatorname{Tr}\sqrt{XX^\dagger}$: the two positive matrices have the same nonzero eigenvalues, namely the squared <singular values>. This follows directly from the <singular value decomposition>.
Set $X=\sqrt\rho\sqrt\sigma$. Then $XX^\dagger=\sqrt\rho\,\sigma\sqrt\rho$ and $X^\dagger X=\sqrt\sigma\,\rho\sqrt\sigma$. Their positive square roots therefore have the same trace, proving
$$
\boxed{F(\rho,\sigma)=\|\sqrt\rho\sqrt\sigma\|_1=F(\sigma,\rho)}.
$$
The argument also covers singular density operators, since zero singular values contribute zero.
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