 Solution

Use the unsquared quantum fidelity convention in the question. For any matrix $X$, its trace norm is $\Vert X{\Vert }_1=\mathrm{Tr}\sqrt{X^{†}X}=\mathrm{Tr}\sqrt{X{X}^{†}}$: 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 $X{X}^{†}=\sqrt{\rho }\sigma \sqrt{\rho }$ and $X^{†}X=\sqrt{\sigma }\rho \sqrt{\sigma }$. Their positive square roots therefore have the same trace, provingThe argument also covers singular density operators, since zero singular values contribute zero.