This is a question from a past exam that I can't seem to figure out. Any tips or hints? Prove $$\forall u,v,x,y,z,w \in \mathbf{R}^+, \frac{u}{v} < \frac{x}{y} \wedge \frac{x}{y} < \frac{z}{w} \implies \frac{u + z}{v+w} < \frac{z}{w}$$
EDIT: My grandfather came up with a different proof than the one Lemur proposed. Here is the formalization of his scratch work that I wrote up:
Let $a$ and $b$ be positive reals such that $a > b$. Then it will follow that $b - a < 0 \iff a > b$. Thus if we can construct two positive reals $n_1$ and $n_2$ such that $n_2 - n_1 < 0$ it will follow that $n_1 > n_2$.
Assume $\displaystyle u,v,x,y,z,w \in \mathbf{R}^+$. Assume $\displaystyle \frac{u}{v} < \frac{x}{y} \wedge \frac{x}{y} < \frac{z}{w}$.
Let $\displaystyle n_1 = \frac{z}{w}$. Let $\displaystyle n_2 = \frac{u+z}{v+w}$.
Then $n_2 - n_1 = \displaystyle \frac{(u+z)w - z(v+w)}{w(v+w)}$ = $\displaystyle \frac{uw+zw-zv-zw}{w(v+w)}$ = $\displaystyle \frac{uw-zv}{w(v+w)}$ = $\frac{\displaystyle \frac{uw-zv}{w}}{\displaystyle \frac{w(v+w)}{w}}$ = $\frac{\displaystyle \frac{uw}{w}-\frac{zv}{w}}{\displaystyle(v+w)}$ = $\frac{\displaystyle u-\frac{zv}{w}}{\displaystyle (v+w)}$ = $\frac{\displaystyle \frac{u}{v}-\frac{z}{w}}{\displaystyle \frac{(v+w)}{v}}$
We know by transitivity that $\frac{u}{v} < \frac{z}{w}$, therefore the numerator is a negative number. We know the denominator is positive since $v$ and $w$ are both positive reals. Thus we have a negative number.
Then ${n_2 - n_1 < 0}$. Then ${n_1 > n_2}$. Then ${\displaystyle \frac{z}{w}} > \frac{u+z}{v+w}$.
Then $\displaystyle \forall u,v,x,y,z,w \in \mathbf{R}^+, \frac{u}{v} < \frac{x}{y} \wedge \frac{x}{y} < \frac{z}{w} \implies \frac{u + z}{v+w} < \frac{z}{w}$
Notice you are given by transitivity
$$ \frac{z}{w} > \frac{u}{v} $$
Hence $zv > uw$. Therefore
$$ z(v + w) = zv + zw > uw + zw = (u+z)w \iff \frac{z}{w} > \frac{u+z}{v+w}$$