Condition number and $LU$ decomposition 
Consider $A: n \times n$ non-singular and the factors $L$ and $U$ of $A$ obtained with partial pivoting strategy, such as: $PA = LU$.
  Proof that $$\kappa_{\infty}(A) \geq \dfrac{||A||_{\infty}}{\min_{j}|u_{jj}|}.$$

The condition number $\kappa_{\infty}(A)$ is defined by $\kappa_{\infty}(A)=||A||_{\infty}||A^{-1}||_{\infty}$.
All I could show was that $\kappa_{\infty}(A) \geq \dfrac{||A||_{\infty}}{n ||U||_{\infty}}$.
But I can't get the "$n$" from the denominator. 
This question seems to me to have some trick that I can´t get it.
I discussed with some colleagues and we were thinking that this question is wrong. But still, we don´t know how to prove it.
 A: Hint:
Let’s drop the permutation matrix (assume $A=LU$) and note that the partial pivoting implies that the absolute values of the elements under the unit diagonal of $L$ are bounded from above by $1$. 
Now try using this:
$$
|u_{ii}^{-1}|=|e_i^TU^{-1}e_i|
=|e_i^TA^{-1}Le_i|
\leq\|A^{-T}e_i\|_1\|Le_i\|_\infty
\leq\|A^{-T}\|_1
=\|A^{-1}\|_\infty.
$$
We used the facts that $|x^Ty|\leq\|x\|_1\|y\|_\infty$ for vectors and $\|X^T\|_1=\|X\|_\infty$ for matrices.
A: Partial answer:
What we're supposed to show is:
$$
||A^{-1}||_{\infty} \geq \frac 1 {\min_{j}|u_{jj}|}
$$
We have further:
$$
||A^{-1}|| = ||(PA)^{-1}||= ||(LU)^{-1}|| = ||U^{-1}L^{-1}||
$$
Now, the diagonal of an inverted triangular matrix  is simply the diagonal of the original triangular matrix with every element inverted.
Let further $i$ be chosen so that $\min_j U_{j,j} = U_{i,i} $, and let $(A)_{k,l}$ depict the entry of $A$ in the $k$-th row and $l$-th column.
Then we have:
$$
||U^{-1}L^{-1}||\ge |(U^{-1}L^{-1})_{i,i}| = \left|\sum_{k=1}^n (R^{-1})_{i,k}(L^{-1})_{k,i}\right|
\\\\
=\left|(R^{-1})_{i,i}(L^{-1})_{i,i}+\sum_{k=1\\ k\neq i}^n (R^{-1})_{i,k}(L^{-1})_{k,i}\right|
\\\\
=\left|\frac 1{\min_j U_{j,j}} + \sum_{k=1\\ k\neq i}^n (R^{-1})_{i,k}(L^{-1})_{k,i}\right|
$$
So, to show the inequality, we now need to show
$$
\sum_{k=1\\ k\neq i}^n (R^{-1})_{i,k}(L^{-1})_{k,i}\ge 0
$$
