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? 
 A: HINT: You can use the idea that $A<B$ if and only if $B-A>0$, as suggested in the comments, or you can simply write all three quantities as fractions with the common denominator $by(y+b)$ and compare numerators. For that comparison you’ll want to use the fact that $\frac{x}y<\frac{a}b$ implies that $bx<ay$.
A: Start here:  observing that for real $a, b, x, y > 0$,  $\frac{x}{y} < \frac{a}{b}$ if and only if $bx < ay$.  Next , add $xy$ to each side, yielding
$bx + xy < ay + xy, \tag{1}$,
or
$x(b + y) < y(a + x), \tag{2}$
or
$\frac{x}{y} < \frac{a + x}{y + b} \tag{3}$
yielding the leftmost inequality.  If we add $ab$ to each side of $bx < ay$, we get
$b(a + x) < a(b + y), \tag{4}$
which with a little re-arranging is
$\frac{a + x}{b + x} < \frac{a}{b}, \tag{5}$
the righrmost inequality.  QED.
And, of course,
Fiat Lux!
A: 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}
A: Another way to think about this, is to view the fractions as proportions.
If I take a mixture with $x\%$ concentration, and mix it with another mixture (of unknown volume) with $y\%$ concentration, what can you say about the concentration of the new mixture?
Must if be between $x\%$ and $y\%$? Why? And if so, how does this prove your question?
