I encountered an inequality when reading a paper. Can someone help to show how to prove it?

Let be the spectral radius of matrix $A$ or $\rho(A)=\max\{|\lambda|, \lambda \text{ are eigenvalues of matrix }A\}$. For matrices $S$ and $T$ with positive spectral radii, and two arbitrary real positive numbers $a$ and $b$, such that $\rho(S) < a < b$ Is the following inequality true? $$b\rho((bI-S)^{-1}T) \leq a\rho((aI-S)^{-1}T)$$ If the above is not true in general, will it be true if $S$ and $T$ are non-negative matrices?

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@Hans: the statement of your question is incomplete. –  Chris Godsil May 5 '13 at 17:21
@Chris Godsil: Thank you for pointing it out. The editor apparently takes the less than symbol "<" followed with no blank as the beginning of an html link and hides all that follows. I have now edited it by inserting spaces after "<". –  Hans May 5 '13 at 17:29

Not true in general, as noted by @SergeiIvanov, but true for (element-wise) nonnegative matrices.

Note that if $\rho(S) < b$, then $b(bI-S)^{-1}=(I-\frac{S}{b})^{-1}=\sum_{i=0}^\infty \frac{S^i}{b^i}$. In particular, thanks to this expansion, if $\rho(S)< a < b$, then $b(bI-S)^{-1}< a(aI-S)^{-1}$ in the componentwise ordering, and thus also $b(bI-S)^{-1}T \leq a(aI-S)^{-1}T$ for any nonnegative $T$. Now, it is a part of the Perron-Frobenius theorem that for any $A,B$ with $0 \leq A \leq B$ then $\rho(A) \leq \rho(B)$, and that's all we need here.

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@Federico Poloni: Thank you for the crisp proof. –  Hans May 6 '13 at 6:35
@Federico Poloni: I posed another related question regarding spectral radius of non-negative matrices mathoverflow.net/questions/129890/a-spectral-radius-inequality. Please take a look. –  Hans May 6 '13 at 20:37
@Federico Poloni: the strict inequality sign in $b\rho((bI-S)^{-1}T)<a\rho((aI-S)^{-1}T)$ is actually incorrect. I have edited the question to reflect that. You proof still goes through with that modification. Would you want to modify it, for prosperity? –  Hans May 15 '13 at 2:37
You are right! One needs an additional hypothesis (irreducibility) to get strict inequalities in P-F. –  Federico Poloni May 15 '13 at 7:18

Not true in general: take $S=T=-I$. Then the inequality boils down to $\frac{b}{b+1}<\frac{a}{a+1}$ which is always false for $b>a>1$.
For positive symmetric matrices, yes. Fix $a$ and let $b\to+\infty$. The l.h.s equals to $\rho((I-\frac1bS)^{-1}T)$ which goes to $\rho(T)$. And the r.h.s. is greater than $\rho(T)$. Indeed, the matrix $S':=(I-\frac1aS)^{-1}$ satisfies $|S'(v)|>|v|$ for all $v\in\mathbb R^n\setminus 0$ (where $n$ is the size of the matrices). Let $v$ be an eigenvector of $T$ corresponding to the maximal eigenvalue $\lambda=\rho(T)$. Then $|S'T(v)|>|T(v)|=\lambda |v|$, hence $\rho(S'T)>\lambda$ by the minimax principle.
@Sergei Ivanov: Nice (partial?) solution. Thank you. But I have two questions. 1. You proved the inequality for positive symmetric matrices with $b\rightarrow\infty$. Is the transition to finite $b>a$ obvious from your present result? 2. Could you please explicate $|S'v|>|v|,\,\forall v\in R^n\\0$? Is this where your symmetry condition on the matrices comes in? –  Hans May 6 '13 at 6:33
@Hans: sorry, I meant "positive definite", not "positive elementwise". The inequality follows by diagonalization. The transition from $b\to\infty$ to a large $b$ is basically the definition of limit. –  Sergei Ivanov May 6 '13 at 10:38
@Sergei Ivanov: I see about positive definite. Regarding $b\rightarrow\infty$, I understand it's the limit. I point is that the limit case does not prove the monotonicity of the spectral radius on finite parameter $a$. For monotonicity, one need to do more. Do you agree? –  Hans May 6 '13 at 13:45
@Hans: yes, sure. I misread the question as "there exist $a$ and $b$ such that...". I am not sure about monotonicity. And actually I now see a flaw in the argument: $S'T$ is not symmetric, so its spectral radius is not equal to the norm. Sorry about this confusion. –  Sergei Ivanov May 6 '13 at 14:19