Suppose $-c<u<c$ and $-c<v<c$. Prove that $\displaystyle -c<\frac{u+v}{1+uv/c^2}<c$.
My attempt:
Adding the two given inequalities, we get $-2c<u+v<2c$.
$|u|<c$ and $|v|<c$
$\implies |u||v|<c^2$
$\implies -c^2<uv<c^2$
$\implies -1<uv/c^2<1$
$\implies 0<1+uv/c^2<2$
$\displaystyle\implies\frac{1}{2}<\frac{1}{1+uv/c^2}$
I think I'm on the right track but I don't know how to proceed.
Consider $$\frac{u+v}{1+uv/c^2}<c$$ Since $|uv/c^2|<1$ the denominator is greater than $0$, so we can multiply by it and then divide by $c$:
$$ \frac uc+\frac vc < 1+\frac uc\frac vc,\\ 1-\frac uc-\frac vc+\frac uc\frac vc > 0,\\ \left(1-\frac uc\right)\left(1-\frac vc\right) > 0. $$
Can you see it now? Can you apply the same logic to $-c<\dots$?