Given two real symmetric matrices $A$ and $B$ of common square size $n$ with no strictly negative eigenvalues, can the symmetric matrix $AB+BA$ have strictly more than $n/2$ eigenvalues which are strictly negative?


The answer to this question is yes, thanks to Junkie. My random examples did not hit a counterexample since I did them in dimension considerably greater than $3$ (typically $8$
or something similar) where it seems quite hard to find counterexamples by taking generic matrices.


This suggests however a series of new questions (which I can unfortunately no longer accept since I
gave already credit to Junkie for a correct answer):


What is the maximal number of strictly negative eigenvalues of $AB+BA$ if $A$ and $B$ are 
definite positive symmetric matrices of common size $n\times n$?


This number is at least roughly $3n/4$ by Junkie's examples (put them along the diagonal). Can it be considerably higher?


I have for example currently no example with $4$ strictly negative eigenvalues for $n=5$.
($3$ strictly negative eigenvalues in dimension $n=5$ are easy to achieve by combining
Junkie's example with an example in dimension $2$ yielding signature $(1,1)$.)


It seems that there is always at least one non-negative eigenvalue (this is obvious if 
$A$ and $B$ have only positive coefficients by Perron-Frobenius and it is probably not very hard in the general case).



A signature inequality?
This [review](http://www.ams.org/mathscinet-getitem?mr=393156) seems to imply that any symmetric real matrix $C$ with positive trace is the Jordan product $(AB+BA)/2$ of two positive definite real matrices $A,B$. If so, then the maximum number of negative eigenvalues of $(AB+BA)/2$ for $n\times n$ symmetric positive definite $A,B$ is $n-1$ (it cannot be $n$ because of the positive definiteness (hence positive trace) of $A^{1/2}BA^{1/2}$, which is conjugate to $AB$).


