proof of arithmetic-geometric-harmonic means inequality
Let be and let be .
We do first the case .
Now we prove the inequality for any power of (that is, for some integer ) by using mathematical induction.
and using the case on the last expression we can state the following inequality
where the last inequality was obtained by applying the induction hypothesis with . Finally, we see that the last expression is equal to and so we have proved the truth of the inequality when the number of terms is a power of two.
3 Inequality for numbers implies inequality for
Finally, we prove that if the inequality holds for any , it must also hold for , and this proposition, combined with the preceding proof for powers of , is enough to prove the inequality for any positive integer.
is known for a given value of (we just proved that it is true for powers of two, as example). Then we can replace with the average of the first numbers. So
On the other hand
which, by hypothesis (the inequality holding for numbers) and the observations made above, leads to:
from where we get that
So far we have proved the inequality between the arithmetic mean and the geometric mean. The geometric-harmonic inequality is easier. Let be .
and so, our proof is completed.
|Title||proof of arithmetic-geometric-harmonic means inequality|
|Date of creation||2013-03-22 12:41:25|
|Last modified on||2013-03-22 12:41:25|
|Last modified by||drini (3)|