## On the Cauchy-Schwartz Inequality

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)
————————————————— .
2a```
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
|uv| < ||u|| ||v|| .
First, we dispose of the (easy) case in which u = 0  — here both sides are obviously just 0 .
Next, given two vectors u0 and v , the formula 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 u0 we can easily infer that the coefficient of t2 below is nonzero too):
p(t) = ||tu + v||2 = (tu + v) • (tu + v) = (uu)t2 + 2(uv)t + vv .
To finish, we use a = uu , b = 2(uv) , and c = vv to translate the b2 < 4ac conclusion to
[2(uv)]2 < 4(uu)(vv) ;
rewriting, dividing both sides by 4, and then taking square roots, we obtain sequentially
4 (uv)2 < 4 ||u||2 ||v||2 ,     (uv)2 < ||u||2 ||v||2 ,     and finally     |(uv)| < ||u|| ||v|| ,
as desired.
 First posted: 2 Apr 2006. Math 221 Home Contents © 2006 by Fred E.J. Linton