Suppose there is a (non-symmetric) real square matrix $A$ with symmetric part $A+A^T$.
What are some conditions on $A$ that are sufficient for $A+A^T$ to be positive definite?
For example, if the eigenvalues of $A$ are strictly positive is $A+A^T$ positive definite? (EDIT: This part of the question is answered in the negative in the comments).
This would then give the result I actually want which is that given two positive definite matrices $C$ and $D$ it follows that the symmetric part of $CD$ is also positive definite. (EDIT: But I think it is still not clear if $CD+DC>0$ – this is (perhaps) a slightly more special case than $A+A^T$ with $A$ having positive eigenvalues.)
Best Answer
To address the problem that the OP actually wanted solved:
This is not true, as can be seen from the following example:
$$C = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}, \quad D = \begin{bmatrix} 100 \\ & 1 \end{bmatrix}.$$
Then their respective spectra are:
$$\sigma(C) = \{1,3\}, \quad \sigma(D) = \{1, 100\}, \quad \sigma(CD+DC) \approx \{-20.2724, 424.272 \}.$$
The reason is simple: when computing $\det(CD+DC)$, the big factor $100$ participates in the positive part only $2$ times, but squared in the negative (counter-diagonal) part.
This might give you some insight how things behave in a more general setting. For example, I'm fairly certain that it can be shown that the following is true: if $C$ and $D$ do not commute, there exists $n \in \mathbb{N}$ such that $CD^n+D^nC$ is not positive definite. All you need to do is observe a $2 \times 2$ submatrix for which $D$ is not a multiply of $I_2$ (WLOG, you can assume that $D$ is diagonal). In other words, any such $D$ can be blown enough to make the above invalid.
This would give you that for every symmetric positive definite $C$ (such that it is not a multiple of identity), you can find a symmetric positive definite $D$ (diagonal, if you want) such that the above does not hold.