Let $A$ and $B$ be $n\times n$ Hermitian positive definite matrices and $k>0$ real. Then $A^k$ is well-defined and experimentally, we have $$\det(A^k+BABA^{-1})\geqslant \det(A^k+BA^{-1}BA),$$or equivalently $$\det(A^k+A^{-1}BAB)\geqslant \det(A^k+ABA^{-1}B).$$ Note that this would imply the inequality $\color{green}{\det (A^2+BAAB)\geqslant \det(A^2+ABBA)}$ found (but presumably unproved) by M. Lin, which is mentioned [in a comment here](https://mathoverflow.net/questions/224121/which-ordering-of-factors-is-needed-to-obtain-this-kind-of-determinantal-inequal#comment553367_224121). 


One striking fact is that $B$ (or $A$, same outcome) can be multiplied by a positive factor, which means that the inequality remains valid for any "proportionality" between the two terms of each sum. Indeed, for any $\lambda>0$, we have equivalently $$\det(A^k+\color{red}\lambda BABA^{-1})\geqslant \det(A^k+\color{red}\lambda BA^{-1}BA).$$



> 
> Any ideas how to prove this conjecture?
> 
> 
> 



A simple but curious determinantal inequality
Let $C = A^{-(k+1)/2} B A B A^{-(k+1)/2}$ and $D = A^{-(k-1)/2} B A^{-1} B A^{-(k-1)/2}$ .


We have to show that $det(I+C) \ge det(I+D)$ .


Now my goal is to apply equation (5.21) in Ando, Majorizations and Inequalities in Matrix Theory, <http://ac.els-cdn.com/0024379594903417/1-s2.0-0024379594903417-main.pdf?_tid=1af72ca2-58c4-11e7-a518-00000aab0f01&acdnat=1498298674_bf9275307496eb46505ba4b5b84a7dcc> :


Choose $E = A^{-(k+1)/2} B A^{-(k+1)/2}$ and $F = A^{k/2+1}$.


Then $C^{1/2} = \vert F E \vert$ and $D^{1/2} = \vert F^{a\_1} E^{b\_1} F^{a\_2} E^{b\_2} \vert$ ,
where $a\_1 = (k/2)/(k/2+1)$, $b\_1 = 1$, $a\_2 = 1/(k/2+1)$, $b\_2 = 0$.


Since $0 \le a\_1 \le b\_1$ and $0 \le a\_2$ and $0 \le b\_2$ and $a\_1 + a\_2 = b\_1 + b\_2 = 1$ it follows that $D^{1/2} \prec\_{\log} C^{1/2}$ and therefore $D \prec\_{\log} C$.


Since $log(1+e^x)$ is convex in $x$ it follows that $I+C$ weakly log majorizes $I+D$ and therefore $det(I+C) \ge det(I+D)$ .


