Arithmetic Mean is greater than geometric mean (2)

 

Theorem

 

Arithmetic mean (A.M.) is greater than geometric mean (G.M.) for three valuables is as follows:

 

 

Equality sign holds if and only if  x = y = z.

 

      This article is dedicated to the proof of the above theorem using different perspectives.

 
 

 

Preliminary

1.        It is assumed that arithmetic mean is greater than geometric mean for two variables is proved :

 

       

 

Equality sign holds if and only if x = y.

 

2.        The reader may find it interesting to investigate the equality case in the inequalities in each

of the following proofs. (We don’t discuss here)

 

3.    If we put   then

                                          (inequality 1)       

       is equivalent to

               ,            (inequality 2)

       Therefore we need to show either (inequality 1) or (inequality 2) in the following proofs.

 

     Method  1

             The easiest proof is :

             (*)

                                                

since a, b, c ≥ 0  and the sum of complete squares in the last factor of (*) must be non-negative.

\            ,  .

 

The weakness of this proof is “how to get (*)?”. We can get (*) by multiplication.

Can we get (*) by factorization ? This is shown at the appendix of this article.

Method  2

 

     Using  A.M  ≥ G.M. for two variables in the preliminary (1) in the above, we get:

                  (i)

                                   (ii)

       \    A.M. of the L.H.S. of (i) and (ii) ≥ G.M. of R.H.S. of (i) and (ii)

                      

                      

                      

                                        \     , .

 

Method  3

 

 

From  , we get

                                                        

     Similarly,                                

                                                      

 

            Adding up the inequalities in the above and divide by 2, we have

                    

 

Multiply both sides by  (a + b + c)  which is non-negative,

the following diagram is helpful in expanding and canceling similar terms :

  

                               

After eliminate the terms, only the red squares remain, we have the inequality:

                                \     , .

 Method  4

 

It is interesting that A.M. ≥ G.M. for four variables can be proved easily :

              
                    

       \

 

       where the proof in fact uses A.M. ≥ G.M. for two variables twice.

Now, we move back to the case for three variables.

This can be done in any one of the following methods :

 

(Method 4a)

 

  Put   in

        

                

                

  \      

 (Method 4b)

 

Put   in

    

    

    

     \       

 

 

Method  5

 

     If  a , b ≥ 0, then   and  a – b  must be of the same sign.

\   

       Expanding and moving terms, we have :

                                                      

       Similarly, we have                      

       And                                       

       Adding the three inequalities, we have:

              

                                      

 

 Method  6

 

For those who are familiar with Calculus the proof below is helpful:

Let   (guess how f(x) is constructed after the proof)

        

        

For turning point,

\  and  

When,

When,

\ f(x) is a minimum point when.

\  

                                

                                

                                

                                

\   

Put x = a, we get    , .

 

 

Method  7

 

Exercise:  Show that   

                 by considering      (1)  

                                                 (2)  

Appendix

       To show that :

               

 

First we note that :

      

               \   

Now,