Consider symmetric positive definite matrices $A_1, \cdots, A_n$ and $B_1,\cdots,B_n$ such that $$A_i^2 < B_i^2$$ for $i \in [n]$, where $A <B$ means that $B-A$ is positive definite.
Is it true that $$\|A_1\cdots A_n\|_F \leq \|B_1\cdots B_n\|_F$$ where $\|A\|_F$ denotes the Frobenius norm of matrix $A$?
According to this answer, the inequality is true for $n=2$.
If it is not true for $n>2$, is there a sufficient condition under which it can be true?
Best Answer
This is not true. When the constraint $A_i^2<B_i^2$ is relaxed to $A_i^2\le B_i^2$, there is a random counterexample: $$ \begin{aligned} &A_1=\pmatrix{0.2144&0\\ 0&0.8922}, \ A_2=\pmatrix{0.3455&-0.2883\\ -0.2883&0.8029}, \ A_3=\pmatrix{0.7817&-0.2633\\ -0.2633&0.5608},\\ &B_1=\pmatrix{0.273970&0.013878\\ 0.013878&0.897168}, \ B_2=\pmatrix{0.4402&-0.2299\\ -0.2299&0.8401}, \ B_3=\pmatrix{0.7818&-0.2635\\ -0.2635&0.5620}\\ \end{aligned} $$ where $$ \frac{\|A_1A_2A_3\|_F}{\|B_1B_2B_3\|_F}-1=0.020842>0. $$ So, if we replace each $B_i$ by $(B_i^2+\epsilon I)^{1/2}$ for a sufficiently small $\epsilon>0$, we obtain a counterexample with $A_i^2<B_i^2$.
Counterexamples like this (subject to the relaxed constraint) can be obtained easily by rejection sampling. Here is my Octave/Matlab script: