Proving $\frac{x+1}{y+1}$ is equal to or less than $\frac{x-1}{y-1}$ I know that $\frac{x+1}{y+1}$ is equal to (when $x = y$) or less than $\frac{x-1}{y-1}$. Suppose we cross multiply, then


*

*$xy - x+ y -1$ is less than or equal to $xy + x - y -1$
So,

*$y- x$ is less than or equal to $x- y$.
But when $y$ is greater than $x$ then the inequality is wrong.
Where am I wrong?


You have
$$
f(x,y)=\frac{x+1}{y+1}-\frac{x-1}{y-1}=
\frac{(xy+y-x-1)-(xy+x-y-1)}{(y^2-1)}=
\frac{2(y-x)}{(y^2-1)}
$$
Then
$$
f(x,y)\begin{cases}
>0 & \text{if $y>x$ and $|y|>1$} \\[4px]
>0 & \text{if $y<x$ and $|y|<1$} \\[4px]
<0 & \text{if $y>x$ and $|y|<1$} \\[4px]
<0 & \text{if $y<x$ and $|y|>1$}
\end{cases}
$$
