How to prove this equation (given two inequalities)? How to prove that:
$$\frac{au+bv}{a+b} < y$$
given that:
$$u<y, v < y$$
Here, a, b are positive integers and u, v and y are real numbers between 0 and 1 (inclusive).

Well the other posters have great answers but you can also see that if $v < y$ then $vb < yb$ since $b > 0$ and we can also see that $ua < ya$ for the same reason. Then we can see that:
\begin{eqnarray}
   ua +vb < ya + yb
\end{eqnarray}
From there its pretty simply algebraic manipulations to get your inequality (Remember you can do a lot of this because $a$ and $b$ are positive, otherwise we would have to reverse the inequality sign)! 
