<s>[INST] Inequality replacement of numbers I'm looking for a contradiction or a proof.
Suppose we have the following inequality:

$a\cdot j \leq b \cdot h$

Where $a,b,j,h \in \mathbb{N}$.
We now replace $a$ with $a_{1}$ and $b$ with $b_{1}$, where $a_{1}, b_{1} \in \mathbb{N}$. We also have the following situation:


*

*$a_{1}\leq a$

*$b_{1}\leq b$

*$b-b_{1}\leq a-a_{1}$


Can we then deduce that $a_{1}\cdot j \leq b_{1} \cdot h$ is true?
For instance:
$4\cdot 6 \leq 7 \cdot 5$
In this case $a = 4, b=7$.
We now replace $a$ with $a_{1} = 3$ so $a - a_{1} = 1$. Since $b-b_{1}\leq a-a_{1} = 1$ we have $b_{1}\in\lbrace 6,7 \rbrace$.
In both cases:


*

*$3\cdot 6 \leq 7 \cdot 5$

*$3\cdot 6 \leq 6 \cdot 5$


The inequality with the replacements of $a,b$ is true.
 [/INST] We cannot, since it is false in general: let $j=1,h=5,a=10,b=2$, so that
$$
a\cdot j=10\cdot 1=10 \leq 10=2\cdot 5=b\cdot h
$$
then let $a_1=9,b_1=1$, thus


*

*$a_1=9\leq 10=a$

*$b_1=1\leq 2=b$

*$b-b_1=2-1=1\leq 1=10-9=a-a_1$


and
$$
a_1\cdot j =9\cdot 1=9 \geq 5=1\cdot 5=b_1\cdot h
$$
</s>