proof of arithmetic-geometric-harmonic means inequality
Let M be max{x1,x2,x3,…,xn} and let m be min{x1,x2,x3,…,xn}.
Then
M=M+M+M+⋯+Mn≥x1+x2+x3+⋯+xnn |
m=nnm=n1m+1m+1m+⋯+1m≤n1x1+1x2+1x3+⋯+1xn |
where all the summations have n terms.
So we have proved in this way the two inequalities at the extremes.
Now we shall prove the inequality between arithmetic mean and geometric mean
.
1 Case n=2
We do first the case n=2.
(√x1-√x2)2 | ≥ | 0 | ||
x1-2√x1x2+x2 | ≥ | 0 | ||
x1+x2 | ≥ | 2√x1x2 | ||
x1+x22 | ≥ | √x1x2 |
2 Case n=2k
Now we prove the inequality for any power of 2 (that is, n=2k for some integer k) by using mathematical induction.
x1+x2+⋯+x2k+x2k+1+⋯+x2k+12k+1 | ||||
= | (x1+x2+⋯+x2k2k)+(x2k+1+x2k+2+⋯+x2k+12k)2 |
and using the case n=2 on the last expression we can state the following inequality
x1+x2+⋯+x2k+x2k+1+⋯+x2k+12k+1 | ||||
≥ | √(x1+x2+⋯+x2k2k)(x2k+1+x2k+2+⋯+x2k+12k) | |||
≥ | √2k√x1x2⋯x2k2k√x2k+1x2k+2⋯x2k+1 |
where the last inequality was obtained by applying the induction hypothesis with n=2k. Finally, we see that the last expression is equal to 2k+1√x1x2x3⋯x2k+1 and so we have proved the truth of the inequality when the number of terms is a power of two.
3 Inequality for n numbers implies inequality for n-1
Finally, we prove that if the inequality holds for any n, it must also hold for n-1, and this proposition, combined with the preceding proof for powers of 2, is enough to prove the inequality for any positive integer.
Suppose that
x1+x2+⋯+xnn≥n√x1x2⋯xn |
is known for a given value of n (we just proved that it is true for powers of two, as example). Then we can replace xn with the average of the first n-1 numbers. So
x1+x2+⋯+xn-1+(x1+x2+⋯+xn-1n-1)n | ||||
= | (n-1)x1+(n-1)x2+⋯+(n-1)xn-1+x1+x2+⋯+xn-1n(n-1) | |||
= | nx1+nx2+⋯+nxn-1n(n-1) | |||
= | x1+x2+⋯+xn-1(n-1) |
On the other hand
n√x1x2⋯xn-1(x1+x2+⋯+xn-1n-1) | ||||
= | n√x1x2⋯xn-1n√x1+x2+⋯+xn-1n-1 |
which, by hypothesis (the inequality holding for n numbers) and the observations made above, leads to:
(x1+x2+⋯+xn-1n-1)n≥(x1x2⋯xn)(x1+x2+⋯+xn-1n-1) |
and so
(x1+x2+⋯+xn-1n-1)n-1≥x1x2⋯xn |
from where we get that
x1+x2+⋯+xn-1n-1≥n-1√x1x2⋯xn. |
So far we have proved the inequality between the arithmetic mean and the geometric mean. The geometric-harmonic inequality is easier. Let ti be 1/xi.
From
t1+t2+⋯+tnn≥n√t1t2t3⋯tn |
we obtain
1x1+1x2+1x3+⋯+1xnn≥n√1x11x21x3⋯1xn |
and therefore
n√x1x2x3⋯xn≥n1x1+1x2+1x3+⋯+1xn |
and so, our proof is completed.
Title | proof of arithmetic-geometric-harmonic means inequality |
Canonical name | ProofOfArithmeticgeometricharmonicMeansInequality |
Date of creation | 2013-03-22 12:41:25 |
Last modified on | 2013-03-22 12:41:25 |
Owner | drini (3) |
Last modified by | drini (3) |
Numerical id | 6 |
Author | drini (3) |
Entry type | Proof |
Classification | msc 26D15 |
Related topic | ArithmeticMean |
Related topic | GeometricMean |
Related topic | HarmonicMean |
Related topic | GeneralMeansInequality |
Related topic | WeightedPowerMean |
Related topic | PowerMean |