Show that $\frac{x-1}{y-1} \leq \frac{x}{y}$

Question

i assume that this is very simple, but i cannot figure out a solution.

can somebody give a proof for $$\frac{x-1}{y-1} \leq \frac{x}{y}$$ given that $$ x \leq y$$ and $$ x \gt 1$$

thank you very much.

Answer 1

Since $y\ge x,x>1$, you have $y-1>0$. So $$ \frac{x-1}{y-1}-\frac{x}{y}=\frac{(x-1)y-x(y-1)}{y(y-1)}=\frac{x-y}{y(y-1)} \le 0$$ namely $$ \frac{x-1}{y-1}\le\frac{x}{y}. $$

Answer 2

\begin{align} x &\le y\\ xy+x &\le xy+y\\ xy-y &\le xy-x\\ y(x-1) &\le x(y-1)\\ \frac {x-1}{y-1} &\le \frac xy \end{align}

Dividing by $y(y-1)$ in the last step maintains the inequality as $y \ge x \gt 1>0$.

Answer 3

$$F=\dfrac xy-\dfrac{x-1}{y-1}=\dfrac{x(y-1)-(x-1)y}{y(y-1)}=\dfrac{y-x}{y(y-1)}$$

What if $F=0?$

$F$ will be $>0$

if even number of terms in $y,y-1,y-x$ are $<0$

If none of those are $<0,y\in[1,0,x]$ or $y>$max$(0,1,x)$

If two of those are $<0,$

If $y<0$ or $y>1,y(y-1)>0\implies y-x>0$

If $0<y<1,y(y-1)<0$ and $y-x<0$

Answer 4

$xy = xy$;

$xy -y \le xy -x$, since $x \le y$

$y(x-1)\le x(y-1)$;

$\dfrac{x-1}{y-1} \le \dfrac{x}{y}$, since $y \ge x >1$.