 Solution

For positive operators, the unique positive square root of an operator respects tensor products: $\sqrt{A\otimes B}=\sqrt{A}\otimes \sqrt{B}$. To verify this, diagonalize $A,B$; the product eigenvectors have eigenvalues $a_i{b}_j$, whose square roots are $\sqrt{a_i}\sqrt{b_j}$. Hence the operator inside the outer square root in the fidelity factors asTaking its positive square root and using $\mathrm{Tr}(C\otimes D)=\mathrm{Tr}C\mathrm{Tr}D$ gives $\boxed{F(\rho \otimes \sigma ,{\rho }^{\prime }\otimes {\sigma }^{\prime })=F(\rho ,{\rho }^{\prime })F(\sigma ,{\sigma }^{\prime })}$.