Let the eigenvalues of $A$ be $a_1\ge a_2\ge \dots \ge a_n$, and the eigenvalues of $B$ be $b_1\ge b_2\ge \dots \ge b_n$. According to the discussion in comments, we are looking for bounds on the singular values of $AB$, denoted $s_1\ge s_2\ge \dots \ge s_n$. The min-max principle is the natural tool to use. Actually, the Wikipedia article does not state the form I'd like to use here: the $k$th largest singular value of $M$ is $\inf\{\|M-T\|:\mathrm{rank}\, T\le k-1\}$. This formula appears in another article with attribution to Allakhverdiev (and no reference). I think the result of Allakhverdiev was the extension to compact operators in Hilbert spaces, but don't know for sure. In general, Wikipedia articles on singular values are a toxic mess.
Claim 1: $s_k\le \min\{ a_i b_{j} : i+j=k+1\}$.
Proof: Let $T$ be an operator of rank $\le i-1$ such that $a_i=\|A-T\|$. Similarly, let $S$ be an operator of rank $\le j-1$ such that $b_{j}=\|B-S\|$. Since $\|(A-T)(B-S)\|\le a_i b_{j}$, it remains to estimate the rank of $(A-T)(B-S)-AB$. Writing $(A-T)(B-S)-AB=-T(B-S)-AS$, we see that the rank is at most $j-1+i-1=k-1$, as desired.
Claim 2: $s_k\ge \max\{ a_i b_{j} : i+j=k+n\}$.
If $A$ and $B$ are invertible, Claim 2 follows by applying Claim 1 to $(AB)^{-1}$. Indeed, the $k$th largest singular value of $AB$ is the reciprocal of the $(n+1-k)$th largest singular value of $(AB)^{-1}$. The $(n+1-k)$th largest singular value of $(AB)^{-1}$ does not exceed $\min \{a_{n+1-i}^{-1}b_{n+1-j}^{-1} : i+j=n+2-k\}$. Decoding these sums of indices, we get Claim 2. The general case follows by considering $A+\epsilon I$ and $B+\epsilon I$, and letting $\epsilon\to 0$.
Edit: Yes, the number of positive or negative eigenvalues in $AB$ and $B$ are the same. The spectrum of $AB$ is identical to the spectrum of $\sqrt{A}B\sqrt{A}$ and by Sylvester's law of inertia, $\sqrt{A}B\sqrt{A}$ and $B$ have the same number of positive/negative/zero eigenvalues.
Best Answer
TravisJ already answered the question for the upper bound on the eigenvalues of the product of two symmetric positive definite matrices, $$\lambda_\text{max}(AB) \le \lambda_\text{max}(A)\lambda_\text{max}(B).$$ To get a lower bound apply the upper bound inequality to the inverse*: $$\lambda_\text{max}\left((AB)^{-1}\right) \le \lambda_\text{max}(A^{-1})\lambda_\text{max}(B^{-1}).$$ Since the eigenvalues of the inverse are the inverse of the eigenvalues, this is the same as, $$\frac{1}{\lambda_\text{min}(AB)} \le \frac{1}{\lambda_\text{min}(A)}\frac{1}{\lambda_\text{min}(B)}.$$ Multiplying through to clear the denominators yields the nice minimum eigenvalue bound, $$\lambda_\text{min}(A)\lambda_\text{min}(B) \le \lambda_\text{min}(AB).$$
*The inverses exist because the matrices are assumed to be positive definite in the question. The result extends to the case where the matrices are positive semi-definite as well. In this case, either $A$ or $B$, are not invertible, and the above proof would divide by zero. However, in this case either $A$, or $B$ has zero as the minimum eigenvalue, and $AB$ has zero as the minimum eigenvalue, so the inequality becomes $0 \le 0$.