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).
 A: Hint: After multiplying through to clear the fraction and rearranging a bit, this is equivalent to asking whether
$$a (y - u) + b (y - v)$$ is positive.
A: Geometrically the point $\alpha x+\beta y$ where $\alpha+\beta=1$ lies in the segment $[x,y]$. Take in your case
$$\alpha=\frac{a}{a+b}\quad\text{and}\quad \beta=\frac{b}{a+b}$$
Can you take it from here?
A: $$au+bv < (a+b)y$$ and thus $$\frac{au+bv}{a+b} < \frac{(a+b)y}{a+b} = y$$
A: 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)! 
A: Note that
$\dfrac{au + bv}{a + b} = \dfrac{a}{a + b} u + \dfrac{b}{a + b} v \tag{1}$
and
$\dfrac{a}{a + b} + \dfrac{b}{a + b} = 1; \tag{2}$
now if $u = v$, (1) and (2) show that
$\dfrac{au + bv}{a + b} = v < y \tag{3}$
by hypothesis; otherwise we can assume $u < v$ in which case (1) and (2) yield
$\dfrac{au + bv}{a + b} = \dfrac{a}{a + b} u + \dfrac{b}{a + b} v < \dfrac{a}{a + b} v + \dfrac{b}{a + b} v  = v < y. \tag{4}$
QED.
Points worth of note:  the restriction $0 < u, v, y < 1$ is not necessary for the result, as long as $u, v < y$;  also the expression (1) shows that $(au + bv) / (a + b)$ is in fact a weighted average of $u, v$ with coefficients $0 < a/a + b, b / a + b < 1$; the assumptions on $a, b$ easily show this to be the case.  Indeed, taking $u < v$, $(au + bv) / (a + b)$ is precisely the real number which is $b / a+ b$ of the way between $u$ and $v$, going in the positive direction, since
$u + \dfrac{b}{a + b}(v - u) = (1 - \dfrac{b}{a + b}) u + \dfrac{b}{a+ b} v = \dfrac{au + bv}{a + b}, \tag{5}$
which provides another way of seeing  $(au + bv) / (a + b) < y$.
Hope this helps.  Happy New Year to One and All,
and as always,
Fiat Lux!!!
A: 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.
