# Writing a proof of an inequality between fractions

I have no idea how to do this.

Suppose $x,y,z,n$ are positive integers. Given that $\frac{x}{y} < \frac{z}{n}$, prove that $$\frac{x}{y} < \frac{x+z}{y+n}$$

Since we have $$\frac xy\lt \frac zn\iff zy-nx\gt 0,$$ we have $$\frac{z}{n}-\frac{x+z}{y+n}=\frac{z(y+n)-n(x+z)}{n(y+n)}=\frac{zy-nx}{n(y+n)}\gt 0$$and $$\frac{x+z}{y+n}-\frac xy=\frac{y(x+z)-x(y+n)}{y(y+n)}=\frac{zy-nx}{y(y+n)}\gt 0.$$
In the first equation, we show $\frac{x+z}{y+n} < \frac{z}{n}$ by showing that their difference is positive.
In the second equation, we show $\frac{x+z}{y+n} > \frac{x}{y}$ again by showing that their difference is positive.
• sorry, bt how do you prove that $(x+z)/(y+n) < z/n$ and $> x/y$ ? Sep 10 '14 at 18:40
• I proved $(z/n)-\{(x+z)/(y+n)\}\gt 0$, which is equivalent to $z/n\gt (x+z)/(y+n).$ Also, I proved $\{(x+z)/(y+n)\}-(x/y)\gt 0$, which is equivalent to $(x+z)/(y+n)\gt x/y$. Hence, I proved that $x/y\lt (x+z)/(y+n)\lt z/n$. Sep 10 '14 at 18:45
• @Aaron: Note that $A\gt B\iff A-B\gt 0$. So, if we prove $A-B\gt 0$, this means that we prove $A\gt B$. Sep 10 '14 at 18:51