Let $ 0 \preceq A \preceq I$ can we relate $Tr( (I-A) B+A C)$ to $Tr(I-A)Tr(B)+Tr(A) Tr(C)$ Let $ 0 \preceq A \preceq I$  and let $B$ and $C$ be two symmetric positive definite matrices. 
Can we related
\begin{align}
{\rm Tr}( (I-A) B+A C)
\end{align}
to
\begin{align}
Tr(I-A)Tr(B)+Tr(A) Tr(C)
\end{align}
via some inequality? 
Note, that in the case $A=aI$  for any $a\in [0,1]$ there is an equality.
Edit:
Note, from a very nice answer below we have:
\begin{align}
\lambda_{\min}(I-A){\rm Tr}(B)+\lambda_{\min}(A){\rm Tr}(C) \le {\rm Tr}( (I-A) B+A C) \le \lambda_{\max}(I-A){\rm Tr}(B)+\lambda_{\max}(A){\rm Tr}(C).
\end{align}
 A: Note that the function 
$$
\langle A,B \rangle = Tr(A^TB)
$$
defines an inner product over the real matrices.  As such, $\|A\|_2^2 = Tr(A^TA)$ is the norm induced by this inner product.  We have
$$
Tr((I-A)B + AC) = Tr((I-A)B) + Tr(AC) =\\
\langle I-A,B \rangle + \langle A,C \rangle \leq\\
\|I - A\|\,\|B\| + \|A\|\,\|C\| = \\
\sqrt{Tr[(I - A)^2]Tr(B^2) + Tr(A^2)Tr(C^2)}
$$
We can do a bit better using the inequality $|\langle A,B \rangle| \leq \sigma_1(A)\sum \sigma_i(B)$. In the symmetric positive semidefinite case, amounts to $|\langle A,B \rangle| \leq \rho(A) Tr(B)$, where $\rho$ denotes the spectral radius.  We now have
$$
\langle I-A,B \rangle + \langle A,C \rangle \leq\\
\rho(I - A)Tr(B) + \rho(A) Tr(C)
$$
This reduces to your result in the case that $A = aI$ for $a \in [0,1]$.

The reverse inequality works perfectly well for our purposes: note that $A \succeq \lambda_{min}(A)I$.  So, $C^{1/2}AC^{1/2} \succeq \lambda_{min}(A)C^{1/2}C^{1/2}$. We therefore have
$$
Tr((I-A)B + AC) = Tr((I-A)B) + Tr(AC) =\\
Tr(B^{1/2}(I-A)B^{1/2}) + Tr(C^{1/2}AC) \geq\\
Tr(\lambda_{min}(I - A)B^{1/2}B^{1/2}) + Tr( \lambda_{min}(A)C^{1/2}C^{1/2}) = \\
\lambda_{min}(I-A)Tr(B) + \lambda_{min}(A) Tr(C)
$$
in retrospect, a similar inequality could have worked in reverse.
