How to prove that $D := ABC$ is also positive definite? Let $A,$ $B$ and $C$ be symmetric, positive definite matrices and suppose that $D := ABC$ is symmetric.
How might I prove that $D$ is also positive definite?
 A: Since $B$ is positive definitie, there is a unique symmetric and positive definite matrix whose square is $B$. Denote this matrix by $B^{\frac{1}{2}}$ and let 
$$\widetilde{A}=B^{\frac{1}{2}}AB^{\frac{1}{2}},\quad \widetilde{C}=B^{\frac{1}{2}}CB^{\frac{1}{2}},\quad \widetilde{D}=B^{\frac{1}{2}}D B^{\frac{1}{2}}.\tag{1}$$
By definition, $\widetilde{A}$, $\widetilde{C}$, $\widetilde{D}$ are symmetric and $\widetilde{A}$, $\widetilde{C}$ are positive definite. Moreover, 
$$\widetilde{D}=\widetilde{A}\widetilde{C}.\tag{2}$$
Similar to $B^{\frac{1}{2}}$, we can define $\widetilde{A}^{\frac{1}{2}}$, and let us denote its inverse by $\widetilde{A}^{-\frac{1}{2}}$. From $(2)$ we know
$$\widetilde{A}^{-\frac{1}{2}}\widetilde{D}\widetilde{A}^{-\frac{1}{2}}=\widetilde{A}^{\frac{1}{2}}\widetilde{C}\widetilde{A}^{-\frac{1}{2}}.\tag{3}$$
Note that the left hand side of $(3)$ is symmetric. Also note that the right hand side of $(3)$ is similar to $\widetilde{C}$, so all its eigenvalues are positive. It follows that the left hand side of $(3)$ is positive definite. As a result, $\widetilde{D}$ is positive definite, and hence $D$ is positive definite.
