Arithmetic and geometric meansIn the following I'll consider sets of positive real numbers
(a1+ ... + aN)/N
whereas their geometric mean is given by
(a1· ... ·aN)1/N
The two quantities always relate in the following manner known as the Arithmetic Mean - Geometric Mean Inequality (AM-GM, for short),
(a1 + ... + aN)/N ≥ (a1· ... ·aN)1/N
Here I am not going to prove the well known inequality but just emphasize a fact that was used by
Cauchy in his proof. Namely, if the inequality holds for all
Thus, assume the inequality holds for all N = 2n and let
ai = (a1 + ... + aN)/N
Since the inequality holds for N = 2n+1 we have
(a1 + ... + a2n+1)/2n+1 ≥ (a1· ... ·a2n+1)1/2n+1
Substituting ai = (a1 + ... + aN)/N for Adding similar terms on the left we get
(N+2n+1-N)(a1 + ... + aN)/(N·2n+1)=
(a1 + ... + aN)/N
which actually says that the arithmetic mean has not been changed by addition of new terms.
(a1 + ... + aN)/N ≥
(a1· ... ·aN)1/2n+1((a1 + ... + aN)/N)(2n+1-N)/2n+1
Dividing by the rightmost term and with one more step to go
((a1 + ... + aN)/N)(1-(2n+1-N)/2n+1) ≥
(a1· ... ·aN)1/2n+1
or
((a1 + ... + aN)/N)N/2n+1 ≥
(a1· ... ·aN)1/2n+1
Now raising both sides to the power of 2n+1/N we finally get
(a1 + ... + aN)/N ≥
(a1· ... ·aN)1/N
Q.E.D.
There is a way to derive a complete proof of the inequality from the Pythagorean Theorem.
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