If $x,y,a,b > 0$ and $\frac{x}{y} < \frac{a}{b}$ then prove $\frac{x}{y} < \frac{x+a}{y+b} < \frac{a}{b}$ So I am not really sure where to start. I understand what I have to do for the proof but I'm thinking that there is someway to use the fact $\frac{x}{y} < \frac{a}{b}$ to deduce which variables are larger than the other. Can anyone help with the starting point? 

Here we are given $\frac{x}{y} < \frac{a}{b}$. Since all values are positive, we get
\begin{align}
 \frac{x}{y}<\frac{a}{b}&\iff\frac{x}{a} < \frac{y}{b}\\
&\iff \frac{x}{a}+1<\frac{y}{b}+1\\
&\iff\frac{x+a}{a}<\frac{y+b}{b}\\
&\iff\frac{x+a}{y+b}<\frac{a}{b}.
\end{align}
Similarly, 
\begin{align}
 \frac{x}{y} < \frac{a}{b}
&\iff\frac{b}{y}<\frac{a}{x}\\
&\iff\frac{b}{y}+1<\frac{a}{x}+1\\
&\iff\frac{y+b}{y}<\frac{x+a}{x} \\
&\iff\frac{x}{y}<\frac{x+a}{y+b}.\\
\end{align}
We have shown that if  $\frac{x}{y} < \frac{a}{b}$ then $\frac{x}{y}<\frac{x+a}{y+b}$ and $\frac{x+a}{y+b}<\frac{a}{b}$ and hence, we have the desired expression 
\begin{equation} 
\frac{x}{y} < \frac{x+a}{y+b} < \frac{a}{b}. 
\end{equation}
