Inequality concerning trace of the product of symmetric positive matrices Let $S$ and $H$ be symmetric positive definite (spd), I have to prove that:
$$ \sqrt[n]{\det{S}}\leq\frac{Tr(SH)}{n\sqrt[n]{\det{H}}}$$

I have already proven that there exists $T$ spd such that $T^2=S$ and that $SH$ is similar to $THT$ and therefore is spd too.
I have also noted that the roots of the determinants are simply the geometric means of the eigenvalues, but have not been able to translate the trace of the product over $n$ into anything meaningful...
I have tried using the inequality of arithmetic and geometric means, but it doesn't really apply since there are two geometric means.
 A: $SH$ is similar to the symmetric positive definite matrix $R=S^{1/2}HS^{1/2}$.
The identity to be proven is equivalent to
$$\sqrt[n]{\det{R}}  \leq \frac{Tr(R)}{n}$$ as $\det{R} = \det{SH} = \det{S}\det{H}$.
$\sqrt[n]{\det{R}}$ is equal to the geometric mean of $R$ eigenvalues (counted with multiplicity). $\frac{Tr(R)}{n}$ is equal to the arithmetic mean of $R$ eigenvalues.
The inequality is therefore a consequence of the fact that geometric mean is less or equal to arithmetic mean.
A: Let $A,B$ be symmetric, positive definite matrices. The main issue in proving this inequality is that the product $AB$ is not necessarily symmetric, as we need the symmetry to conclude that (??? not sure here why symmetry is needed). In any case, consider the matrix $AB (AB)^\dagger$ where the $\dagger$ denotes the Hermitian conjugate. Now, we have that:
$$
\mathrm{Tr}(AB(AB)^\dagger) = \mathrm{Tr}(ABB^\dagger A^\dagger) = \mathrm{Tr}(ABBA)=  \mathrm{Tr}(AABB) = \mathrm{Tr}((AB)^2)
$$
with the last few manipulations coming from the fact that $A,B$ are Hermitian, and that the trace is invariant under cyclic permutation. Now, $(AB)^2$ is positive definite and Hermitian, so by the spectral theorem, the square root of $AB$ is also symmetric and also positive definite. Applying standard AM-GM argument + the multiplicativity of the determinant nets us the inequality we need. 
