Lower bound for the trace of product of two symmetric matrices i am stuck on finding a lower bound of $tr(XY)$ of two symmetric matrices in $M_{n}(\mathbb{R})$.
I know that it holds $tr(XY)=tr(YX)$ and thus $tr(XY-YX)=0$ and i can remember, that XY-YX is also symmetric. I know that it holds $tr(XY)\leq \lambda(X)^{T}\cdot \lambda{Y}$, where $\lambda{A}$ is the sorted vector of eigenvalues of the matrix $A$. Can you give me a hint for finding a lower bound?
 A: There does not exist $k>0$ s.t., for every $X,Y>0$, $tr(XY)\geq k||X||||Y||$.
Proof. Take, when $n=2$, $X=diag(1,\epsilon),Y=diag(\epsilon,1)$ where $\epsilon$ tends to $0_+$.
A: If XY is positive definite  and a is the smallest eigenvalue then na is a lower bound on the trace.  If XY is positive semi -definite with rank r, and and a is the smallest non-zero eigenvalue, then ra is a lower bound on the trace.
If XY is negative definite  or semi-negative definite, then you can similarly find an upper bound B on the trace of (-XY). Then -B will be a lower bound on trace(XY).
If XY has mixed positive and negative eigenvalues, the smallest eigenvalue of XY is a lower bound.  Without further conditions on X and Y, I do not see anything better (although someone else might). 
Connecting any lower bounds to the eigenvalues of X and Y seems difficult.  Here are some examples: Let X be a diagonal matrix with 1/2's on the diagonal.  If Y is a diagonal matrix with all 1/2's on the diagonal, then the eigenvalues of XY are all 1/4 and the trace is n/4.  But if Z is a diagonal matrix with all 10's on the diagonal, the trace of XZ is 5n.
You can try to cast some of the above arguments about positive definite matrices in terms of the product of the smallest eigenvalues of X and Y, but I think that lower bound is worse than the one I suggested above. 
