By similar arguments as for the proof of the goldenthompson inequality (see "Log majorization and complementary GoldenThompson type inequalities" by T.Ando and F.Hiai) we can show that for all A,B symmetric positive definite we have $$\\log(A)+\log(B)\_{tr}\leq \\log(A^{1/2}BA^{1/2})\_{tr},$$ where $\log$ is the matrix logarithm and $\\cdot\_{tr}$ is the trace norm, i.e. $\A\_{tr}=\sqrt{tr(AA^*)}$. My conjecture is that for each $n\in \mathbb{N}$ there exists a constant $c_n$ such that $$\\log(A)+\log(B)\_{tr}\geq c_n\\log(A^{1/2}BA^{1/2})\_{tr}$$ for all symmetric positive definite matrices $A,B$ of dimension $n$. However I have no idea how to prove it. The question is interesting because it would show that the two metrics $d_1$, $d_2$ defined by $$d_1(A,B)=\\log(A)\log(B)\ \text{ and } d_2(A,B)=\\log(A^{1/2}BA^{1/2})\$$ on the space of positive definite Matrices are strongly equivalent.
Check out:
Reverse inequality to Golden–Thompson type inequalities: Comparison of $e^{A+B}$ and $e^Ae^B$ JeanChristophe Bourin, Yuki Seo (2007), Linear Algebra and its Applications Volume 426, Issues 2–3, 15 October 2007, Pages 312316.

$\begingroup$ Thanks for this reference. But I dont understand why the inequalities in this reference should be sharp. $\endgroup$ – user35593 Nov 24 '13 at 9:10

$\begingroup$ Now I found a simple counterexample to my conjecture for the case $n=2$: $$A_m=\exp\left(\begin{matrix}m & 0\\0 & 0\end{matrix}\right) \text{ and } B_m=\exp\left(\begin{matrix}m & 1\\1 & 0\end{matrix}\right)$$ $\endgroup$ – user35593 Dec 12 '13 at 13:10