Whare $\dagger$ stands for the Moore-Penrose generalized inverse and $0 \leq \lambda_1(B) \leq \ldots \leq \lambda_n(B)$ stand for the eigenvalues of $B \in \mathbb R^{n \times n}$.
The proof for the same inequality when $B$ is invertible goes as follows.
For $A,B \in \mathbb R^{n \times n}$ both positive definite matrices. First you show that for $D = B^{-1/2}AB^{-1/2}$ \begin{equation} > \operatorname{Tr}((A + B)^{-1}A ) = \operatorname{Tr}( (D+I)^{-1}D ). \quad \quad \quad(1) > \end{equation} Since $(D+I)^{-1}D = I - (D+I)^{-1}$ it follows that $0 < (D+I)^{-1}D < I$.
Since $(D+I)^{-1}D$ and $D$ have the same eigenvectors and it's easy to see that $(D+I)^{-1}D < D$.
Now, if $v_1 \ldots v_n$ form an orthonormal basis of eigenvectors of $B$ with $0 < \lambda_1 \leq \ldots \leq \lambda_n$ their associated eigenvalues. we get \begin{align*} \operatorname{Tr}((A + B)^{-1}A ) & = > \sum_{i=1}^{n} v^t(D+I)^{-1}D v \\ > & \leq k + \sum_{i=k+1}^{n} v^t D v \\ > & = k + \sum_{i=k+1}^{n} \frac{1}{\lambda_i} v^t A v \\ > & \leq k + \frac{1}{\lambda_{k+1}} \operatorname{Tr}(A) \end{align*}
I am having trouble to reproduce this idea when $B$ is not invertible. To prove (1) you first note that
\begin{align*} (A + B)^{-1}A & = \left(B^{1/2}(B^{-1/2}A B^{-1/2} + I)B^{1/2}\right)^{-1}A \\ & = B^{-1/2}(B^{-1/2}AB^{-1/2} + I)^{-1}B^{-1/2} A \end{align*} I cannot do this for the Moore-Penrose inverse. If $\operatorname{Im}(A)$ is the space spanned by the columns of A, then $$(AB)^{\dagger} = B^\dagger A^{\dagger} \iff \operatorname{Im}( A^*A B ) \subseteq \operatorname{Im}(B) \text{ and } \operatorname{Im}( B^*B A ) \subseteq \operatorname{Im}(A) $$ Even if the previous condition hold it there is a problem because $A \neq (B^{1/2})^{\dagger} B^{1/2} A B^{1/2}(B^{1/2})^{\dagger}$.
Obviously the inequality in the question only makes sense when $\lambda_{k+1}(B) > 0$.