We have two real-valued Positive Semi-definite matrices, $A$ and $B$, both $n$ by $n$.
Can the minimum eigenvalue $\lambda_\mathrm{min}(AB)$ be lower-bounded using the following eigenvalue sequences?
$\lambda_1(A)\ge \lambda_2(A) \dots \lambda_\mathrm{min}(A)$
$\lambda_1(B)\ge \lambda_2(B) \dots \lambda_\mathrm{min}(B)$
Not much are known other than PSDness of two matrices, except rank of $B$ is $n-1$.
The minimum eigenvalue is necessarily $0$. We can prove that this is the case as follows:
We begin by noting that the product $AB$ has non-negative eigenvalues. We can see this by noting that if $A$ is invertible, then $AB$ is similar to the positive semidefinite matrix $$ A^{1/2}BA^{1/2} = A^{-1/2}(AB)A^{1/2} $$ If $A$ fails to be invertible, then it suffices to note that $AB = \lim_{\epsilon \to 0^+}(A+\epsilon I)B$.
Because $B$ is singular, the product $AB$ is singular.
So, $AB$ is a singular matrix with non-negative eigenvalues. So, its minimum eigenvalue must be $0$.