TITLE: Sum of two velocities is smaller than the speed of light
QUESTION [14 upvotes]: 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.

REPLY [6 votes]: 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
$$ 1+xy-(x+y)=(1-x)(1-y)>0.
$$