We assume everyone knows that a real polynomial p of degree 2 , of the form
p(t) = ax2 + bx + c , with a ≠ 0 ,
has a graph taking on one of the following approximate shapes: ∩ , ∪ ,
and that to find the roots of such a polynomial, i.e., the values of x
at which the graph of p crosses the x-axis, one calculates the (real)
values given by the well-known quadratic formula
–b ± √(b2 – 4ac)
Of course the graph of p may need not ever actually cross the x-axis,
perhaps because it never touches it at all, or perhaps because it touches it at just
one point; these situations come about exactly when the expression inside the radical
sign, that is, the expression b2 – 4ac , is negative
or zero, i.e., b2 – 4ac < 0 , or what amounts
to the same thing,
b2 < 4ac .
This is the case, in particular, when p(t) > 0 for all real t .
It is the above observation that serves us best as we seek to establish the Cauchy-Schwartz
Inequality, which asserts (of two vectors u and v , whether both column
vectors in Rn or both row vectors in Rn ) that
|u • v| < ||u|| ||v|| .
First, we dispose of the (easy) case in which
u = 0 —
here both sides are obviously just 0 .
Next, given two vectors u ≠ 0 and v ,
p(t) = ||tu + v||2 pretty obviously defines a
real-valued function p for which it’s also pretty obvious that
p(t) > 0 for all real t .
We now evaluate p(t) to discover that p
is a polynomial of degree 2 (not less, because from u ≠ 0 we can
easily infer that the coefficient of t2 below is nonzero too):
p(t) = ||tu + v||2 =
(tu + v) • (tu + v) =
(u • u)t2 +
2(u • v)t + v • v .
To finish, we use a = u • u ,
b = 2(u • v) , and c = v • v
to translate the b2 < 4ac conclusion
to [2(u • v)]2 <
4(u • u)(v • v) ;
rewriting, dividing both sides by 4, and then taking square roots, we obtain sequentially
4 (u • v)2 <
4 ||u||2 ||v||2 ,
(u • v)2 <
||u||2 ||v||2 , and finally
|(u • v)| < ||u|| ||v|| ,