Transformation that preserves an increasing ratio between vectors Consider two vectors $x=(x_1,x_2,\ldots,x_n)$, $y= (y_1,y_2,\ldots,y_n)$ such that all $x_i,y_i>0$ and 
\begin{align}
\frac{y_1}{x_1}\le \frac{y_2}{x_2}\le\cdots\le \frac{y_n}{x_n}
\end{align}
Now consider an upper triangular matrix $A$ with elements $a_{ij}\ge 0$
I want to prove (or come up with conditions on $A$ so) that 
\begin{align}
\frac{[yA]_1}{[xA]_1}\le \frac{[yA]_2}{[xA]_2}\le\cdots\le \frac{[yA]_n}{[xA]_n}
\end{align}
For example, for $n=3$,
\begin{align}
A = \left( \begin{array}{ccc}
a_{11} & a_{12} & a_{13} \\
0 & a_{22} & a_{23} \\
0 & 0 & a_{33} \end{array} \right)
\end{align}
We would want to show
\begin{align}
\frac{a_{11}y_1}{a_{11}x_1}\le \frac{a_{12}y_1 + a_{22}y_2}{a_{12}x_1+a_{22}x_2}\le \frac{a_{13}y_1 + a_{23}y_2 + a_{33}y_3}{a_{13}x_1 + a_{23}x_2 + a_{33}x_3}
\end{align}
I am interested in proving this for general $n$.

Update: I'm thinking that a sufficient condition for this to hold is that the column sums are increasing in the column index. That is, for $n=3$, $a_{11} \le a_{12} + a_{22} \le a_{13}+a_{23}+a_{33}$. Is this true?
 A: Necessarily, (*) each column of $A$ must be a non-zero vector.
The conditions when $n\leq 3$ are simple. When $n=2$, each $\geq 0$ matrix satisfying (*) answers the question. When $n=3$, and $A=\begin{pmatrix}a&b&c\\0&d&e\\0&0&f\end{pmatrix}\geq 0$, the NS condition is $\det(\begin{pmatrix}b&c\\d&e\end{pmatrix})\geq 0$. I think that we can generalize this result.
EDIT. Proposition. The required condition is: for every $i,j,l$ satisfying $i<l\leq j$, one  has $\det(\begin{pmatrix}a_{i,j}&a_{i,j+1}\\a_{lj}&a_{l,j+1}\end{pmatrix})\geq 0$.
Outline of the proof. Let $C_i$ be the $i^{th}$ column of $A$, $(e_j)_j$ be the canonical basis of $\mathbb{R}^n$. The reasoning is by recurrence. Thus, it suffices to show that $y^TC_{n-1}/x^TC_{n-1}\leq y^TC_{n}/x^TC_{n}$, that is  $f(x,y)=y^TC_{n-1}x^TC_n-x^TC_{n-1}y^TC_n\leq 0$. Note that $C_n=D+a_{nn}e_n$ where $d_n=0$; in the sequel, $C$ denotes $C_{n-1}$.
Thus $f(x,y)=(y^TCx^TD-x^TCy^TD)+a_{nn}(y^TCx_n-x^TCy_n)$. The key is that the first part does not depend on $x_n,y_n$ and the second one is always $\leq 0$. Finally, it remains to find the ad hoc condition ensuring that $g(x,y)=y^TCx^TD-x^TCy^TD\leq 0$. One has $g(x,y)=y^T(CD^T-DC^T)x$; the matrix $CD^T-DC^T$ is skew symmetric and its entries above the diagonal are the $\det(.)$ cited in the proposition. After, it's easy. You can convince yourself by expanding $g(.)$ with hand...
