An inequality related to matrix norm, inverse matrix The question is that:
We have two matrices $A,B\in \mathbf{C}^{n\times n}$, A is nonsingular and B is singular, let $||\cdot ||$ be $\textbf{any}$ matrix norm, prove
$||A-B||\geq 1/||A^{-1}||$.
My idea is 
$||(A-B)||\cdot||A^{-1}||\geq||A^{-1}(A-B)||=||I-A^{-1}B||$
And I am confused here. How to calculate the right side?
$\color{red}{Note:}$ the original question may be wrong when the norm doesn't have sub-multiplicative property. See the comments below. 
Thanks for everybody's help!
 A: HINT:
$B= A+ (B-A) = A\cdot( I - A^{-1}(A-B) ) = A \cdot (I - C)$
where $C = A^{-1}(A-B)$.  Assume that $||A-B||< ||A^{-1}||^{-1}$. Then $||C||\le ||A^{-1} || \cdot ||A-B|| < 1$. Now use the following fact:
If $C$ is a matrix with norm $||C||<1$ then the series $\sum_{n\ge 0} C^n$ is absolutely convergent and its sum is the inverse of $I-C$. In particular: $||C||<1$ implies $(I-C)$ invertible. 
Use the above to conclude $B$ is invertible, contradiction.
A: Since $B$ is singular there exists some $v\neq 0$ with $\|v\|=1$  such that $Bv=0$. Thus,
$$\|A-B\|_2^2\geq \|(A-B)v\|^2=\|Av\|^2=v^*A^*Av\geq \lambda_{min}(A^*A)\\ =\lambda_{min}\left[(A^{-1}(A^{-1})^*)^{-1}\right]=\frac{1}{\lambda_{max}(A^{-1}(A^{-1})^*)}=\frac{1}{\|A^{-1}\|_2^2}$$
EDIT: For a general induced $p$-norm ($p>1$): Similarly there exists some $v\neq 0$ with $\|v\|_p=1$  such that $Bv=0$. Thus, for the induced matrix $p$-norm
$$\|A-B\|_p\geq \|(A-B)v\|_p=\|Av\|_p\qquad\qquad(1)$$
Define now $q>0$ such that $\frac{1}{p}+\frac{1}{q}=1$. Then using Holder inequality
$$\|Av\|_p\|A^{-1}\|_q\geq \|Av\|_p\frac{\|A^{-1}v\|_q}{\|v\|_q}\geq \|v\|_2^2\|v\|_{q}^{-1}\qquad (2)$$
Thus for a general $p$-norm ($p>1$) it holds true from (1), (2) that
$$\|A-B\|_p\geq \min_{v\in Ker(B), \|v\|_p=1} \left\{\|v\|_2^2\|v\|^{-1}_q\right\}\frac{1}{\|A^{-1}\|_q}$$
