Posting the solution from the website
The proof is this is quite tricky. We let x be a scalar and note from property 1
that
0 < (xu + v)
. (xu +
v)
= u . u x2
+ 2 u . v x + v
. v
Since the dot product produces a scalar, the above equation is a
quadratic in x. A
quadratic that is always positive has nonpositive
discriminant. Hence
(2 u . v)2 - 4(u
. u)(v . v)
<
0
dividing by 4 gives
(u . v)2 - (u
. u)(v . v)
<
0
or
(u . v)2 < (u
. u)(v .
v)
Taking the square root of both sides produces the result.