Thanks for your efforts, everything is clear now. It seems that your bound is smaller than mine.

Thanks for your comments, it's clear now. Regarding your reply 1): I don't know $\lambda_{\min}$ or $\lambda_{\max}$ can be interpreted in this form. 3): I tried and arrived at $\frac{\lambda_{\max}(B)}{\lambda_{\max}(B)+\lambda_{\min}(A-B)} - \frac{\lambda_{\max}(A)-\lambda_{\min}(A-B)}{\lambda_{\max}(A)}=\frac{\lambda_{\min}(A-B)(\lambda_{\min}(A-B)-\lambda_{\max}(A)+\lambda_{\max}(B))}{\lambda_{\max}(A)\lambda_{\max}(B)+\lambda_{\max}(A)\lambda_{\min}(A-B)}$. Since it's hard to decide whether $(\lambda_{\min}(A-B)-\lambda_{\max}(A)+\lambda_{\max}(B))>0$ or not, I stop here.

I couldn't understand well, 1) $\lambda_m$ refers which matrix? 2) Does ''minimal of maximal'' mean ''minimal to maximal'' ? 3) Is your answer implying my answer, that is, $\alpha = \frac{\lambda_{\max}(B)}{\lambda_{\max}(B)+\lambda_{\min}(A-B)}$ can infer to $\alpha=1-\lambda=\frac{\lambda_{\max}(A)-\lambda_{\min}(A-B)}{\lambda_{\max}(A)}$?

Can $\alpha =\frac{M(B)}{M(B)+m(A-B)}$ such that $\alpha A\geq B$?

Thanks for your efforts, I have 2 questions regarding your answer: 1) why let $A$ and $B$ be $n×n$ matrices and $S\subset R^n$ be the unit sphere? 2) $(\alpha Ax, x)-(Bx,x)>0$ may not imply $m(\alpha A-B)>0$.

Thanks for you suggestion, after following your suggestion, I arrange it as follows: $A-B>\epsilon I$ which is equivalent to $A-\epsilon I> B$, also $(I-\epsilon A^{-1}) A> B$, but this still can't show $\alpha A>B$ where $0< \alpha <1$.

If you mean my proposition is always true?

$A>B$ means $A-B$ is positive-definite matrix.