On what condition $trace(A) \ge trace(AB)$? Considering $A$ and $B$ are positive semidefinite real symmetric matrices, on what conditions we can have $trace(A) \ge trace(AB)$?
 A: Under the extra assumption that $A,B$ commute: Then $A,B$ are orthogonally diagonalizable simultaneously.
Let $P$ be an orthogonal matrix, and $D_1,D_2$ be diagonal such that 
$$A=PD_1P^T \\
B=PD_2P^{T}$$
Then 
$$AB=PD_1D_2P^{T}$$
Therefore, if $\lambda_1,.., \lambda_n$ are the eigenvalues of $A$ and $\beta_1,.., \beta_n$ are the eigenvalues of $B$, in any order, there exists a permutation $\sigma$ such that 
$$\tr(AB)=\lambda_1 \beta_{\sigma(1)}+...+\lambda_n \beta_{\sigma(n)}$$
The question you are asking in this case is basically the following
Question Given some non-negative numbers $\lambda_1,.., \lambda_n$  and $\beta_1,.., \beta_n$, (and a permutation $\sigma$ which can actually be ignored by reordering $\beta$) under which condition do we have
$$\lambda_1 \beta_{\sigma(1)}+...+\lambda_n \beta_{\sigma(n)}\leq \lambda_1+..+\lambda_n ?$$
Note that for any such $\lambda, \beta$'s you can take $A,B$ to be the diagonal matrix with those numbers on the diagonal for an example/counterexample.
If $A$ is not the zero matrix, then the most obvious answer is : all eigenvalues of $B$ are $\leq 1$.
If $A,B$ don't commute, the question seems to be way to complicated.
