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Let $A, B, C$ be symmetric positive definite matrices, define $R(A,B)=\max\{\rho(A^{-1}B), \rho(B^{-1}A) \}$, where $\rho(\cdot)$ means the spectral radius. Is it obvious $R(A,C)\le R(A,B)R(B,C)$? How to prove this?

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Yes. First, recall that $$ \rho(AB) \;\leq\; \rho(A)\rho(B) $$ for any symmetric matrices $A$ and $B$. Then \begin{align*} R(A,B) R(B,C) \;&=\; \max\{\rho(A^{-1}B),\rho(B^{-1}A)\} \max\{\rho(B^{-1}C),\rho(C^{-1}B)\} \\ &\geq\; \max\{\rho(A^{-1}B)\rho(B^{-1}C),\rho(B^{-1}A)\rho(C^{-1}B)\} \\ &\geq\; \max\{\rho(A^{-1}C),\rho(C^{-1}A)\} \\ &=\; R(A,C). \end{align*}

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