Sum of two velocities is smaller than the speed of light Using the Lorentz transformation from special relativity, we get that the sum of two velocities can be expressed as
$$u=\frac{u'+v}{1+\frac{u'v}{c^2}}.$$
Given that $|u'|,|v| \le c$, I want to prove that $|u| \le c$, ie. that the velocity never exceeds $c$. However I am struggling to produce this bound. I have tried to bound the denominator from above but this produces zero and have tried a case wise approach but this has got me no where either.
 A: Setting $x=u'/c, y = v/c, z = u/c$ as in another answer, i.e. expressing velocities as fractions of the speed of light, you want to show
$$ -1 < \frac{x+y}{1+xy} < 1 \quad\text{for $-1< x,y < 1$}.
$$
Note that the denominator is positive, so you want to show
$$ -1-xy < x+y < 1+xy.$$
The first inequality follows from
$$ x+y+1+xy = (1+x)(1+y) > 0,
$$
the second from
$$ 1+xy-(x+y)=(1-x)(1-y)>0.
$$
A: Let $u'=c-x$ and $v=c-y$ with $x\geq0$ and $y\geq0$. Then we have:
$$
u=\frac{(c-x)+(c-y)}{1+\frac{(c-x)(c-y)}{c^2}}=\frac{2c-x-y}{2-\frac{x}{c}-\frac{y}{c}+\frac{xy}{c^2}}=c\frac{(2-\frac{x}{c}-\frac{y}{c})}{(2-\frac{x}{c}-\frac{y}{c})+\frac{xy}{c^2}}\leq c
$$
(since $\frac{xy}{c^2} \geq 0$)
You can similarly set $u'=-c+x$, $v=-c+y$ to show that $u\geq-c$.
A: 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 $c$.
A: 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.
A: 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.
A: Hint: If you're familiar with techniques in optimization, try and choose values of $u'$ and $v$ so as to optimize $u$. If you can prove that the optimal values for making $u$ large yield a value of $u$ at or approaching $c$, then you are done.
