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).

Here is a generalization that you may find useful.
Suppose that $w_1, w_2 \cdots w_n>0$ and $x_1, x_2, \cdots x_n$ are any numbers. Then
if all $x_1$ are greater than $m$ and less than $M$ i.e.
$$
m < x_1, m < x_2, \cdots m <x_n$$
and
$$ x_1<M, x_2<M, \cdots x_n < M$$
Then
$$
m < \frac{w_1 x_1 + w_2 x_2 \cdots +w_n x_n} {w_1 + w_2 \cdots+ w_n} < M
$$
Proof: First consider
$$
\frac{w_1 (x_1-m) + w_2 (x_2-m) \cdots + w_n (x_n-m)} {w_1 + w_2 \cdots +w_n}$$
Each term on the numerator and denominator is positive. So the ratio is positive.
Hence
$$
0<\frac{w_1 (x_1-m) + w_2 (x_2-m) \cdots + w_n (x_n-m)} {w_1 + w_2 \cdots +w_n}
= \\
\frac{w_1 x_1 + w_2 x_2 \cdots +w_n x_n} {w_1 + w_2 \cdots+ w_n} 
-
\frac{w_1 m + w_2 m \cdots +w_n m} {w_1 + w_2 \cdots+ w_n} 
$$
Factor out $m$ from the second sum on the right to get
$$
0<
\frac{w_1 x_1 + w_2 x_2 \cdots +w_n x_n} {w_1 + w_2 \cdots+ w_n} 
-
m
$$
The other inequality can be obtained by considering
$$
\frac{w_1 (M-x_1) + w_2 (M-x_2) \cdots + w_n (M-x_n)} {w_1 + w_2 \cdots +w_n}$$
Note:
$$
\frac{w_1 x_1 + w_2 x_2 \cdots +w_n x_n} {w_1 + w_2 \cdots+ w_n} 
$$
is called the weighted average, $w_1$, $w_2$ etc. are called the weights. The result basically says that the average has to be between the min and max of the numbers whose average you are calculating.
