Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Cauchy-Schwarz Inequality


Date: 11/13/1999 at 23:36:15
From: Nick Krupar
Subject: Cauchy-Schwarz Inequality

Prove: 
       n                    n              n
     [sum(a_j)(b_j)]^2 <= (sum[j(a_j)^2])(sum[[(b_j)^2]/j])
      j=1                  j=1            j=1

I've shown that, where <...> is the inner-product: 
 
       n
     [sum(a_j)(b_j)]^2 = <(a_1, ..., a_n),(b_1, ..., b_n)>^2
      j=1 

          n             n 
     <= (sum[(a_j)^2])(sum[(b_j)^2])
         j=1           j=1

I know that:

      n               n
     sum[(a_j)^2] <= sum[j(a_j)^2]
     j=1             j=1

      n               n
     sum[(b_j)^2] >= sum[[(b_j)^2]/j]
     j=1             j=1


Date: 11/14/1999 at 08:17:31
From: Doctor Anthony
Subject: Re: Cauchy-Schwarz Inequality

The Cauchy-Schwarz inequality is that for real numbers ai, bi then

  (a1.b1 + a2.b2 + .... + an.bn)^2 <= (a1^2 + a2^2 + ... + an^2)*
                                        (b1^2 + b2^2 + ... + bn^2)

Proof:

For any x we have

   0 <= (a1.x+b1)^2 + (a2.x+b2)^2 + ... + (an.x + bn)^2

      = (a1^2 + ... + an^2)x^2 + 2(a1.b1 + ... + an.bn)x +
                                                  (b1^2 + ... + bn^2)
      =  Ax^2 + 2Bx + C

since y = Ax^2 + 2Bx + C >= 0 for ALL x it follows that the equation 
Ax^2 + 2Bx + C = 0 cannot have two distinct roots. And so

   (2B)^2 <= 4AC    (discriminant of the quadratic equation <= 0)
so
    B^2 <= AC
so
   (a1.b1 + ... + an.bn)^2 <= (a1^2 + ... + an^2)(b1^2 + ... + bn^2)

which is the required inequality.

A further inequality that is often quoted is:

   (x1 + x2 + x3 + ...+ xn)^2 <= n(x1^2 + x2^2 + ... + xn^2)

If we take just 3 values, we require to show

   (x1 + x2 + x3)^2 <= 3(x1^2 + x2^2 + x3^2)

   x1^2+x2^2+x3^2+2x1.x2+2x2.x3+2x3.x1 <= 3(x1^2 + x2^2 + x3^2)

and so

   2x1.x2 + 2x2.x3 + 2x3.x1 <= 2x1^2 + 2x2^2 + 2x3^2

   0 <= x1^2 - 2x1.x2 + x^2 + x1^2 - 2x1.x3 + x3^2 + x2^2 - 2x2.x3
                                                              + x3^2
   0 <= (x1-x2)^2 + (x1-x3)^2 + (x2-x3)^2

which is clearly true since each term on the right is >= 0

The same pattern is true if you take higher values of n. You will 
always be able to put the inequality into the form shown above.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Number Theory

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/