0$, can we always find a positive real number $\alpha$, $0<\alpha < 1$ such that $\alpha A \geq B $?

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when toggle format	what		by	license	comment
Jun 8, 2015 at 5:27	comment	added	Alex Ravsky		I expect that we can similarly prove that $\alpha A\ge B$ provided $\alpha=\frac{M(B)}{m(A-B)+M(B)}$.
Jun 8, 2015 at 5:25	comment	added	Alex Ravsky		2) The sphere $S$ is compact, so a continuous positive-valued function on $S$ attains its (positive) infimum (that is why I wrote $”\min”$ instead of $”\inf”$.)
Jun 8, 2015 at 5:25	comment	added	Alex Ravsky		1) I just fixed $n$. $S$ represents all non-zero vectors of the space $\Bbb R^n$. I considered only a real case, but I hope that the proof for the complex case is similar.
Jun 7, 2015 at 12:44	comment	added	wayne		Can $\alpha =\frac{M(B)}{M(B)+m(A-B)}$ such that $\alpha A\geq B$?
Jun 7, 2015 at 8:24	vote	accept	wayne
Jun 7, 2015 at 8:14	comment	added	Włodzimierz Holsztyński		The general definition considers the complex field, not real.
Jun 7, 2015 at 7:49	comment	added	wayne		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 $m(\alpha A-B)>0$.
Jun 7, 2015 at 5:17	history	answered	Alex Ravsky	CC BY-SA 3.0