If $A>B>0$, can we always find a positive real number $\alpha$, $0<\alpha < 1$ such that $\alpha A \geq B $? Suppose we have positive-definite matrices $A$, $B$,   if $A>B>0$, can we always find a positive real number $\alpha$,  $0<\alpha < 1$ such that $ \alpha A \geq B $? If it has, then what is it?
 A: It seems the following.
Let $A$ and $B$ be $n\times n$ matrices and $S\subset\Bbb R^n$ be the unit sphere. Then positive definiteness of the matrix $A$ is equivalent to 
$$m(A)=\min\{(Ax,x):x\in S\}>0.$$
Similarly, we  have 
$$m(A-B)=\min\{(A-B)x,x):x\in S\}>0.$$
Put 
$$M(B)=\max\{(Bx,x):x\in S\}>0.$$
Let $\alpha>\frac{M(B)}{m(A-B)+M(B)}<1$ be any number and $x\in S$ be any vector. Then 
$$(\alpha Ax,x)-(Bx,x)=$$ $$\alpha((A-B)x,x)-(1-\alpha)(Bx,x)\ge$$ $$\alpha m(A-B)-(1-\alpha)M(B)=$$ $$\alpha(m(A-B)+M(B))-M(B)>$$ $$\frac{M(B)}{m(A-B)+M(B)} (m(A-B)+M(B))-M(B)>0.$$ 
Then 
$$m(\alpha A-B)= \min\{(\alpha A-B)x,x):x\in S\}>0.$$
So the matrix $\alpha A−B$ is positive-definite, that is $\alpha A>B$
A: The @Alex Ravsky's answer is good, however, I still want to share my answer:
Proof: To prove $\alpha A\geq B$ where $0<\alpha<1$, we introduce an extra parameter $\lambda$, $0<\lambda<1$, $\alpha=1-\lambda$, such that $(1-\lambda)A\geq B$, which is equivalent to find a $\lambda$, $0<\lambda<1$, such that  $(1-\lambda)A\geq B$.
Arrange it, we get, $\lambda A\leq A-B$, to make it hold, we shall find $\lambda$ such that $\lambda A \leq \lambda_{\min}(A-B)I$ holds, which is equivalent to  $\lambda I \leq \lambda_{\min}(A-B) A^{-1}$. Further, let $\lambda = \frac{\lambda_{\min}(A-B)}{\lambda_{\max}(A)}$, the above equations are automatically satisfied.   
