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).

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!!!
