Sum of two velocities is smaller than the speed of light
With only basic analysis, you can do it by fixing one of the two variables.
You can prove that, for a fixed $u'<c$:
- The function $f_{u'}(v)=\frac{u'+v}{1+\frac{u'}{c^2}v}$ is a monotonically increasing function.
- $f_{u'}(c)=c$
From $1$, you can conclude that if $v<c$, $f_{u'}(v)<f_{u'}(c)$,and including $2$, that gives you $f_{u'}(v)<c$.
This proves that if $u',v$ are both smaller than $c$, that $\frac{u'+v}{1+\frac{u'v}{c^2}}$ will also be smaller than $c$.
Such relativistic sum of speeds (here denoted as $\oplus$) can be written in terms of a simple pullback: $$ u\oplus v\stackrel{\text{def}}{=}c\cdot\tanh\left(\text{arctanh}\tfrac{u}{c}+\text{arctanh}\tfrac{v}{c}\right)$$ since $\tanh$ is an increasing function with range $(-1,1)$, $\left|u\oplus v\right|< c$ immediately follows.
Here's an elementary proof which avoids transcendental functions.
Letting $x=u'/c, y = v/c, z = u/c$ the equation for the Einstein sum becomes
$$z = \frac{x+y}{1+x y}\tag{1}$$
and we have to prove that
$$-1\lt z \lt 1\tag{2}$$
for $-1\lt x \lt 1, -1 \lt y \lt 1$.
Now we let
$$x\to \frac{1-r}{1+r},y\to \frac{1-s}{1+s}\tag{3a}$$
or
$$r \to \frac{1-x}{1+x}, s\to \frac{1-y}{1+y}\tag{3b}$$
which transforms $x\in (-1,+1)$ to $r \in (\infty, 0)$ and $y\in (-1,+1)$ to $s \in (\infty, 0)$ .
In other words, we have parametrized $x$ and $y$ with positive parameters $r$ and $s$.
Substituting (3) in (1) gives
$$z = \frac{1-t}{1+t}\tag{4}$$
with $t = r s$. Hence we have $t \in (\infty, 0)$ and from (4) follows (2). QED.