How to prove the following relation about the minimum of two ratios?

Question

I want to know the proof of following relation $$\min\left(\frac{x}{x+c},\frac{y}{y+c}\right)=\frac{\min(x,y)}{\min(x+c,y+c)}$$ where $c$ is some positive constant and $x,y$ can have only non-negative values. Any help will be much appreciated. Thank you.

Answer 1

If $x<y$, then $x + c < y + c$, so $min(x,y) = x$ and $min(x+c,y+c) = x + c$

Also: $xc<yc$

So:

$$\frac{x}{x+c} =$$

$$\frac{x(y+c)}{(x+c)(y+c)} =$$

$$\frac{xy+xc}{(x+c)(y+c)} < \frac{xy+yc}{(x+c)(y+c)} =$$

$$ \frac{y(x+c)}{(x+c)(y+c)} = $$

$$\frac{y}{y+c}$$

So:

$$min(\frac{x}{x+c},\frac{y}{y+c}) = \frac{x}{x+c} = \frac{min(x,y)}{min(x+c,y+c)}$$

Likewise, if $y<x$, then

$$min(\frac{x}{x+c},\frac{y}{y+c}) = \frac{y}{y+c} = \frac{min(x,y)}{min(x+c,y+c)}$$

And if $x=y$ then obviously the result holds as well.

Answer 2

Without loss of generality, assume that $x<y$.

The ratio $\frac{x}{x+c}$ will always be less than $\frac{y}{y+c}$

Reason: $\frac{x}{x+c}$ = $1 - \frac{c}{x+c}$ and $\frac{y}{y+c}$ = $1 - \frac{c}{y+c}$

$\frac{c}{x+c} > \frac{c}{y+c}$ because $x+c < y+c$

This gives us:

$min(\frac{x}{x+c}, \frac{y}{y+c}) = \frac{x}{x+c} = \frac{min(x,\ y)}{min(x+c,\ y+c)}$

QED