Prove that the A.M. of two numbers is always greater than its G.M.
Answers
Answer:
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,
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