Let $A$ be a diagonalizable $n\times n$ matrix with eigenvalues $\lambda_1,\dots, \lambda_n$, $B$ an $n\times n$ matrix, and $\lambda$ an eigenvalue of $A+B$. Show that $$\min\limits_{j=1,\dots,n}|\lambda-\lambda_j|\le ||C||_p||C^{-1}||_p||B||_p$$ where $C$ is a nonsingular matrix such that $C^{-1}AC$ is diagonal and $p=1,2,\infty$.
I'm having difficulty figuring out where to start. If given some guidance I'm sure I can easily get the rest. I know that under the assumption $A$ is diagonalizable gives $C^{-1}AC=diag(\lambda_1,\dots,\lambda_n)$, but I am failing to see how I will use the other assumptions. Any input would be greatly appreciated!
Here's an argument for the $p=\infty$ case.
Suppose that $(A+B)v=\lambda v$. Since $C^{-1}AC$ is diagonal, we have $C^{-1}ACe_j=\lambda_je_j$, where $e_1,\ldots,e_n$ are the canonical basis. So $Av_j=\lambda_jv_j$, where $v_j=Ce_j$.
Assume $A$ is invertible (when $A$ is not invertible, we may tweak the eigenvalues slightly so that it is, and the we approximate the estimates). Then $v_1,\ldots,v_n$ are a basis. Write $v=\sum_jr_jv_j$ for some coefficients $r_j$. Now \begin{align} Bv&=\lambda v-Av=\sum_j r_j \lambda v_j-\sum_jr_j Av_j=\sum_jr_j(\lambda-\lambda_j)v_j\\ \ \\ &=\sum_jr_j(\lambda-\lambda_j)Ce_j. \end{align} Thus $$\tag1 \sum_j r_j(\lambda-\lambda_j)e_j=C^{-1}Bv. $$ Since $v\ne0$, $r_k=\max\{|r_j|:\ j\}\ne0$. Then $$\tag2 |r_k|\,|\lambda-\lambda_k|\leq\left\|\sum_jr_j(\lambda-\lambda_j)e_j\right\|_\infty=\|C^{-1}Bv\| $$ Also, since $v=\sum_jr_jv_j=\sum_jr_jCe_j$, we have that $$\tag3 \|v\|=\left\| C\,\sum_jr_je_j\right\|\leq\|C\|\,\left\| \sum_jr_je_j\right\| _\infty=\|C\|\,|r_k| $$ From $(2)$ and $(3)$, $$ |\lambda-\lambda_k|\leq\frac1{|r_k|}\,\|C^{-1}Bv\|\leq\frac1{|r_k|}\,\|C^{-1}\|\,\|B\|\,\|v\|\leq\|C^{-1}\|\,\|B\|\,\|C\|. $$