proof of arithmetic-geometric-harmonic
means inequality∗
drini†
2013-03-21 14:11:44
Let M be max{x1 , x2 , x3 , . . . , xn } and let m be min{x1 , x2 , x3 , . . . , xn }.
Then
M + M + M + ··· + M
x1 + x2 + x3 + · · · + xn
≥
n
n
n
n
n
m= n = 1
1
1
1 ≤ 1
1
1
1
m
m + m + m + ··· + m
x1 + x2 + x3 + · · · + xn
M=
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
√
x1 − 2 x1 x2 + x2
x1 + x2
x1 + x2
2
2
≥
0
≥
0
√
2 x1 x2
√
x1 x2
≥
≥
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.
∗ hProofOfArithmeticgeometricharmonicMeansInequalityi
created: h2013-03-21i
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1
x1 + x2 + · · · + x2k + x2k +1 + · · · + x2k+1
2k+1
=
x1 +x2 +···+x2k
2k
+
x2k +1 +x2k +2 +···+x2k+1
2k
2
and using the case n = 2 on the last expression we can state the following
inequality
x1 + x2 + · · · + x2k + x2k +1 + · · · + x2k+1
2k+1
s
x1 + x2 + · · · + x2k
x2k +1 + x2k +2 + · · · + x2k+1
≥
2k
2k
q
√
√
2k
x1 x2 · · · x2k 2k x2k +1 x2k +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 x1 x2 x3 · · · 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 + · · · + xn
≥ n x1 x2 · · · xn
n
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
+···+xn−1
x1 + x2 + · · · + xn−1 + x1 +x2n−1
=
=
=
n
(n − 1)x1 + (n − 1)x2 + · · · + (n − 1)xn−1 + x1 + x2 + · · · + xn−1
n(n − 1)
nx1 + nx2 + · · · + nxn−1
n(n − 1)
x1 + x2 + · · · + xn−1
(n − 1)
2
On the other hand
s
n
x1 + x2 + · · · + xn−1
x1 x2 · · · xn−1
n−1
r
x1 + x2 + · · · + xn−1
√
= n x1 x2 · · · xn−1 n
n−1
which, by hypothesis (the inequality holding for n numbers) and the observations
made above, leads to:
n
x1 + x2 + · · · + xn−1
x1 + x2 + · · · + xn−1
≥ (x1 x2 · · · xn )
n−1
n−1
and so
x1 + x2 + · · · + xn−1
n−1
n−1
≥ x1 x2 · · · xn
from where we get that
x1 + x2 + · · · + xn−1
≥
n−1
√
n−1
x1 x2 · · · 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 + · · · + tn
≥ n t1 t2 t3 · · · tn
n
we obtain
r
1
1
1
1
1 1 1
1
x1 + x2 + x3 + · · · + xn
≥ n
···
n
x1 x2 x3
xn
and therefore
√
n
x1 x2 x3 · · · xn ≥
n
1
x1
+
and so, our proof is completed.
3
1
x2
+
1
x3
+ ··· +
1
xn