Cauchy-Schwarz Inequality
|u*v| ≤ ||u|| ||v||, such that u and v are non-zero vectors.
Proof:
If |u*v| = ||u|| ||v||, then u and v are parallel vectors, i.e.: u = kv, where k is any scalar in real number.
Let p(t) = ||tv - u||^2 ≥ 0
p(t) = (tv - u)*(tv - u) = v*v t^2 - 2t(u*v) + u*u
Let a = v*v, b = 2u*v, c = u*u
p(t) = at^2 - bt + c
p(b/2a) = a(b/2a)^2 - b(b/2a) + c = b^2 /4a - b^2 /2a + c = -b^2 /4a + c ≥ 0
∴ c ≥ b^2 /4a; 4ac ≥ b^2
With the original values:
4(v*v)(u*u) ≥ (2u*v)^2
(v*v)(u*u) ≥ (u*v)^2
||v||^2 ||u||^2 ≥ (u*v)^2
∴|u*v| ≤ ||u|| ||v|| QED