The two root-fidelity trace expressions agree

Put $A=\sqrt{\rho }\sqrt{\sigma }$. Since $A^{\ast }=\sqrt{\sigma }\sqrt{\rho }$, we have $A^{\ast }A=\sqrt{\sigma }\rho \sqrt{\sigma }$ and $A{A}^{\ast }=\sqrt{\rho }\sigma \sqrt{\rho }$. Both are non-negative. By [L1], the eigenvalues of $A^{\ast }A$ and $A{A}^{\ast }$ are the squares of the singular values of $A$; because $A$ is an endomorphism, the SVD lists the same $d$ singular values on both sides, with zero multiplicity included.

The unique non-negative square roots of $A^{\ast }A$ and $A{A}^{\ast }$ therefore have the same eigenvalue list, namely the singular values of $A$. Their traces both equal the sum of those singular values, which is $\mathrm{\parallel }A{\mathrm{\parallel }}_1$ by [L1]. Substituting $A{A}^{\ast }=\sqrt{\rho }\sigma \sqrt{\rho }$ proves the stated identity and, by [F1], its equivalent root-fidelity expression.