Does increasing both variables increase $\frac{y}{x-y}$?

Question

I have this fraction with positive variables, x and y: $$\frac{y}{x-y}$$ $$x-y<1$$

Does increasing x and y always increase the whole fraction? I think it is true because any number divided by a number between 0 and 1 will make it bigger. But my concern is that as we increase y, x-y will change too.

Answer 1

Let $x_0=1, y_0 = 0.5$, then $x_0-y_0 = 0.5 < 1$, then $\frac{y_0}{x_0-y_0}=1$

Let's make $x$ grows faster than $y$ and yet satisfying $x_1 - y_1 < 1$.

Let $x_1 = 1.5, y_1 = 0.6$, then $x_1-y_1=0.9 < 1$, $\frac{y_1}{x_1-y_1}=\frac{0.6}{0.9}=\frac23 < 1$

Answer 2

You need to define how you change $x$ and $y$. If you change them by an equal amount, so $x$ becomes $x+a$ and $y$ becomes $y+a$, then it is true: $$\frac{y+a}{(x+a)-(y+a)}=\frac{y+a}{x-y}=\frac{y}{x-y}+\frac{a}{x-y}>\frac{y}{x-y}$$

Answer 3

Note that for $a\ge 0$

• numerator increase $y \to y+a$
• $x-y$ is constant since $(x+a)-(y+a)=x-y$

therefore for $x-y>0$

$$\frac{y+a}{(x+a)-(y+a)}=\frac{y+a}{x-y}\ge \frac{y}{x-y}$$

otherwise, as shown in other answers, not always the inequality holds and we need to consider case by case.

In a more general approach we can consider the gradient of $f(x,y)$

$$\nabla f(x,y)=(f_x,f_y)$$

and for a given direction $\vec n$ study the conditions for which

$$\nabla f(x,y)\cdot n >0$$

Answer 4

You say $x$ and $y$ are positive. I'm going to assume $x-y > 0$ as well so $0 < x- y < 1$ which means $y < x < y + 1$

If $y' > y$ and $x' > x$ and we still have $y' < x' < y' + 1$, whether $\frac {y'}{x' - y'} > \frac {y}{x - y}$ or not will depend on the rate of at which $y$ increases compares with how $x -y$ increases/decreases or stays steady.

$\frac {y'}{x' - y'} > \frac {y}{x - y} \iff$

$(x - y)y' > (x'-y')y \iff$

$xy' > x'y$.

We can take this further. $y' > y$ so let $y' = y + e$ for some $e$. If $x' = x + e$ i.e. they both increase by the same ammount.

We get $x(y + e) > (x+e)y \iff$

$xe > ye$ which is true. (And this is also true because $x' - y' = (x+e) - (y+e) = x - y$ so the denominater stays the same while $y$ increases).

However if the $y' = y + d$ and $x' = x + e$ we get

$xy' > x'y\iff$

$x(y + d) > (x + e)y \iff$

$xd > ey \iff$

$\frac xy > \frac ed$

Which means it could decrease if the proportion of how much $x$ increases compared to how much $y$ increases is greater than the proportion of $x$ to $y$.

So for example if $x:: y$ is $3::2$ say $x= .3$ and $y= .2$ but $e::d$ is $2::1$ so $e = .2$ and $d = .1$ we get that

$\frac {.2}{.3 - .2} > \frac {.3}{.5 -.3}$ we actually get a decrease.

But if $x::$ is $3::2$ say $x = .6$ and $y= .4$ ane $e::d$ is $4::3$ say $.4$ and $.3$ we will get an increase:

$\frac {.4}{.6-.4} < \frac {.8}{.9 - .8}$.