proof of arithmetic-geometric-harmonic means inequality
For the Arithmetic Geometric Inequality, I claim it is enough to prove that if
∏ni=1xi=1 with xi≥0 then ∑ni=1xi≥n. The arithmetic geometric inequality for y1,…,yn will follow by taking
xi=yin√∏nk=1yk. The geometric harmonic inequality follows from the arithmetic geometric by taking xi=1yi.
So, we show that if ∏ni=1xi=1 with xi≥0 then ∑ni=1xi≥n by induction on n.
Clear for n=1.
Induction Step: By reordering indices we may assume the xi are increasing, so xn≥1≥x1. Assuming the statement is true for n-1, we have x2+⋯+xn-1+x1xn≥n-1. Then,
n∑i=1xi≥n-1+xn+x1-x1xn |
by adding x1+xn to both sides and subtracting x1xn. And so,
n∑i=1xi | ≥n+(xn-1)+(x1-x1xn) | ||
=n+(xn-1)-x1(xn-1) | |||
=n+(xn-1)(1-x1) | |||
≥n |
The last line follows since xn≥1≥x1.
Title | proof of arithmetic-geometric-harmonic means inequality |
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Canonical name | ProofOfArithmeticgeometricharmonicMeansInequality |
Date of creation | 2013-03-22 15:09:37 |
Last modified on | 2013-03-22 15:09:37 |
Owner | Mathprof (13753) |
Last modified by | Mathprof (13753) |
Numerical id | 10 |
Author | Mathprof (13753) |
Entry type | Proof |
Classification | msc 26D15 |