Cauchy-Schwartz Inequality

 

Motivation:

 

                           Previously,  we have the following equality for the area of triangle:

                                                

                                    OAB  = ½

 

                           But can we always say

 

                                       is larger than 0 ?

 

                           Although we did not talk about  n-dimensional vector,  what will it be

                           about the relationship between  and   if

 

                                               = (a₁, a_2,  a₃, … , a_n)

                                               = (b₁, b₂, b₃, …, b_n )

 

                           Actually, this is very famous theorem known as Cauchy-Schwartz Inequality.

 

 

 

 

 

 

Theorem ( Cauchy-Schwartz Inequality )

                  a₁, a_2,  a₃, … , a_n    R  ,   b₁, b₂, b₃, …, b_n  R .  Then

 

       (a₁² + a₂² + a₃² + …  + a_n² )(  b₁² + b₂² + b₃² + … + b_n² )   (a₁b₁ + a₂b₂ + … + a_nb_n )²

 

                 The equality holds when

                                        =  = … =

 

 

 

 

Proof:

 

              In general,  we have

                                 (ax-b)²  0      if  a, b  R  for any x  R .

 

              So, let’s consider the following:

                                (a₁x – b₁)²  0

                                (a₂x – b₂)²  0

                                 …

                                (a_nx – b_n)²  0

 

                               (a₁x – b₁)² + (a₂x – b₂)² + … + (a_nx – b_n)²    0

 

               This inequality always holds for any x  R.  Then

 

           (a₁² + a₂² + … + a_n² ) x² – 2(a₁b₁ + a₂b₂ + … + a_nb_n ) x + (b₁² + b₂² + … + b_n² )  0

 

 

                      For a quadratic form

                               Ax² + Bx + C  0  for any x  R,

 

                      if A   0,  then we have

                                     B² – 4 AC   0 .

 

                       The equality holds when  x =   ,  the minimum value

                                                 = 0

 

                      This is the criteria that the quadratic form has no distinct real roots.

 

                                                   

 

                     So,  we apply the result on the problem here:

 

              (2(a₁b₁ + a₂b₂ + … + a_nb_n ))²  - 4 (a₁² + a₂² + … + a_n² ) (b₁² + b₂² + … + b_n² )  0

        (a₁b₁ + a₂b₂ + … + a_nb_n )²   (a₁² + a₂² + … + a_n² ) (b₁² + b₂² + … + b_n² )

 

               When does the equality hold ?   The equality holds when  the equality of every of the following holds simultaneously:

 

                                (a₁x – b₁)²  0

                                (a₂x – b₂)²  0

                                 …

                                (a_nx – b_n)²  0

 

                So,

                               x =  =  = … =

 

 

 

Theorem:

                  For two vectors  and ,  we have

 

                                       0

 

                 The equality holds when the two vectors are in parallel, in other words,

                                                   = t  ,  t  R .