Let $A$, $B$ be two square $n\times n$ real matrices and suppose that $B=B^\top$ is positive definite. Denote by $\langle\cdot,\cdot\rangle$ the trace inner product in the space of symmetric matrices, that is $\langle X,Y\rangle:=\mathrm{tr}(X^\top Y)$, for $X,Y$ symmetric matrices.
My question: Suppose that $$ \langle A+A^\top,\alpha(AB+BA^\top)+\beta B\rangle=0, $$ for all $\alpha$, $\beta$ real numbers. Does the previous condition implies $A+A^\top=0$?
A simple remark. If $AB=BA$ then it is easy to see that the answer is affirmative (just pick $\alpha=1$, $\beta=0$).
Let me use a notation for the trace product that is much easier to type $$X:Y = {\rm tr}(X^TY) = {\rm tr}(XY^T)$$ If $B$ is symmetric, then the term
$$\eqalign{ C &= AB + (AB)^T = AB + B^TA^T = AB + BA^T \cr }$$ is symmetric.
Next note that the linear combination of two symmetric matrices, i.e. $$\eqalign{ S &= \alpha C + \beta B \cr }$$ is itself symmetric.
So we have this expression
$$\eqalign{ 0 &= (A+A^T):S \cr &= A:S + A^T:S \cr &= A:S + A:S^T \cr &= 2A:S \cr 0 &= A:S \cr }$$ and that final equality means that $A$ is indeed skew-symmetric.