$A\in\mathbb{R}^{n\times n}$ is a non-singular (invertible) matrix and we denote its $p$-norm with $||A||_p$ which is:
$$||A||_p = \underset{x\neq 0}{\text{max}} \dfrac{||Ax||_p}{||x||_p}$$
which $x\in \mathbb{R}^n$. Also $K_p(A)$ is its condition number equal to $||A||_p||A^{-1}||_p$. Now assume we add matrix $E$ to $A$ such that $A+E$ is a singular (not-invertible) matrix. Show $$K_p(A)\geq \dfrac{||A||_p}{||E||_p}$$
This is what I tried:
$$\begin{align}
K_p(A)&\geq \dfrac{||A||_p}{||E||_p}\\||A||_p||A^{-1}||_p&\geq \dfrac{||A||_p}{||E||_p}
\end{align}$$
Here, if $||A||_p$ is zero, we are done. Else:
$$\begin{align}
||A^{-1}||_p&\geq \dfrac{1}{||E||_p}\\
||A^{-1}||_p||E||_p&\geq 1
\end{align}$$
Here's where I'm stuck. I also know: $$\begin{align} 1=||I||_p=||AA^{-1}||_p\leq ||A||_p||A^{-1}||_p \end{align}$$
But I can't go further to prove what question is asking for. I don't know how to connect that $A+E$ is singular to these inequalities.
Any help or hint is immensely valuable to me!
Note that for $x\in N(A+E)$ nonzero, where $N(A+E)$ is the nullspace of $A+E$ (remember that $A+E$ is singular) we have $$ 0 = (A+E)x\implies Ax=-Ex\implies x=-A^{-1}Ex $$ Thus $$ 1=\frac{\|A^{-1}Ex\|_p}{\|x\|_p}\leq\|A^{-1}E\|_p\leq\|A^{-1}\|_p\|E\|_p. $$