I am given that the Bures distance between two density matrices (i.e. semi positive-definite self-adjoint maps which have trace equal to unity) $\rho_1,\rho_2$ is: $D_B(\rho_1,\rho_2) = 2(1-\|\rho_1^{1/2} \rho_{2}^{1/2}\|)$ where $\|\cdot\|$ is the $L^1$-norm defined by $\|A\| = \mathrm{Tr}(\sqrt{A^\star A})$,
I try to prove that $D_B(\rho_1,\rho_2) =0 \implies \rho_1 = \rho_2$.
I have already proven that $0\leq D_B(\rho_1,\rho_2)\leq 2$. My attempt is to write using that $\mathrm{Tr}(\rho_1) = \mathrm{Tr}(\rho_2) = 1$:
$$1 = \|\rho_1^{1/2}\rho_{2}^{1/2}\| = \|\rho_1\| = \|\rho_2\|,$$
however does it follow from this that $\rho_1 = \rho_2$ should hold?
Conceptually this is very simple -- just write this out in a particularly nice way and apply Cauchy Schwarz.
It suffices the show $\|\rho_1^{1/2} \rho_{2}^{1/2}\|=\|\rho_1^{1/2} \rho_{2}^{1/2}\|_{S_1}\leq 1$ with equality iff $\rho_{1}=\rho_{2}$. I include the $S_1$ since this is a Schatten 1 norm.
It is also convenient to quasi-linearize the Schatten 1 norm by writing it as
$\Big \Vert A \Big \Vert_{S_1}=\max_U \Big \vert \text{trace}\big(U^*A\big)\Big\vert$ with $U\in U_n$. This comes the from Polar Decomposition of $A$ (with optional use (i) of von Neumann Trace Inequality to verify the maximum or (ii) diagonalize the HPSD matrix in A's Polar Decomposition and apply triangle inequality when writing out the trace).
Proof
$\|\rho_1^{1/2} \rho_{2}^{1/2}\|_{S_1}$
$=\max_U \Big \vert \text{trace}\Big(U^*\rho_1^\frac{1}{2} \rho_2^\frac{1}{2}\Big) \Big \vert $
$=\max_U \Big \vert \text{trace}\Big(\big(U^*\rho_1^\frac{1}{2}\big) \rho_2^\frac{1}{2}\Big) \Big \vert $
$=\max_U \Big \vert \text{trace}\Big(\big(\rho_2^\frac{1}{2}\big)\big(U^*\rho_1^\frac{1}{2}\big)\Big) \Big \vert $
$=\max_U \Big \vert \text{trace}\Big(\big(\rho_2^\frac{1}{2}\big)^*\big(U^*\rho_1^\frac{1}{2}\big)\Big) \Big \vert $
$\leq\Big \Vert \rho_2^\frac{1}{2}\Big \Vert_F\cdot \Big \Vert U^*\rho_1^\frac{1}{2}\Big \Vert_F$
$= \text{trace}\big(\rho_2\big)^\frac{1}{2}\cdot \text{trace}\big(\rho_1\big)^\frac{1}{2}$
$= 1 \cdot 1$
$=1$
with equality iff $\rho_1 = \rho_2$
further explanation:
examining the equality conditions of Cauchy Schwarz, which imply $\eta \cdot U^*\rho_1^\frac{1}{2} = \rho_2^\frac{1}{2}$ $1=\vert \eta\vert^2\cdot \Big\Vert U^*\rho_1^\frac{1}{2}\Big \Vert_F^2=\vert \eta\vert^2\cdot\text{trace}\big(\rho_1\big)= \text{trace}\big(\rho_2\big)=1\implies \eta$ is on the unit circle.
Thus we can say
$Q\rho_1^\frac{1}{2} = \big(\eta \cdot U^*\big)\rho_1^\frac{1}{2} = \rho_2^\frac{1}{2}$
$\implies \rho_1^\frac{1}{2}Q^* = \rho_2^\frac{1}{2}$
$\implies \rho_1=\big(\rho_1^\frac{1}{2}Q^*\big)\big(Q\rho_1^\frac{1}{2}\big)=\big(\rho_1^\frac{1}{2}Q^*\big)\rho_2^\frac{1}{2}=\rho_2^\frac{1}{2}\rho_2^\frac{1}{2} =\rho_2$