Algebra+ 07 - Averages: UNIZOR.COM - Math+ & Problems - Algebra

Notes to a video lecture on http://www.unizor.com

Algebra+ 07

Note
We will use both terms average and mean interchangeably, but usually with proper qualification, like geometric mean or harmonic average.

Problem A

Given that a fence around a rectangular field should have the length L, what should the sides of this field be to maximize the field's area?
What is the maximum area of a field in this case?

Hint
Arithmetic mean (average) of two positive real numbers is not less than their geometric mean (average).
These averages are equal only if the participating numbers are equal.

Answer
All sides must be equal to L/4 to form a square field.
The area will then be L²/16.


Problem B

In lecture UNIZOR.COM - "Math 4 Teens" - "Math Concepts" - "Induction" - "Averages" we have proven that arithmetic mean of N positive real numbers is greater or equal to their geometric mean:
(X₁+X₂+...+X_N)/N ≥
≥ (X₁·X₂·...·X_N)^1/N
Prove now that geometric mean of N positive numbers is greater or equal to their harmonic mean:
(X₁·X₂·...·X_N)^1/N ≥
≥ N/(1/X₁+1/X₂+...+1/X_N)

Hint B
Use the theorem about arithmetic and geometric means mentioned above for numbers 1/X₁, 1/X₂ etc.


Problem C

Prove that quadratic mean of N positive numbers is greater or equal to their arithmetic mean:
sqrt[(X₁²+X₂²+...+X_N²)/N] ≥
≥ (X₁+X₂+...+X_N)/N
where sqrt is a square root function.

Hint C
The proof in general case is similar to the one with only a few numbers.
Let's analyze the situation with only 2 components.
Start from what's necessary to prove
sqrt[(a²+b²)/2] ≥ (a+b)/2
Since all numbers are positive, get rid of sqrt function by raising both sides of an inequality to the power of 2 and apply invariant transformations.
(a²+b²)/2 ≥ (a+b)²/4
2·(a²+b²) ≥ a²+2a·b+b²
a²+b²−2a·b ≥ 0
(a−b)² ≥ 0
which is obvious, and all transformations are reversible.
Therefore, the proof is to derive the required inequality by reversed transformations from the obviously correct last inequality.

Let's illustrate the same approach for 4 numbers.
sqrt[(a²+b²+c²+d²)/4] ≥
≥ (a+b+c+d)/4
Since all numbers are positive, get rid of sqrt function by raising both sides of an inequality to the power of 2.
(a²+b²+c²+d²)/4 ≥
≥ (a+b+c+d)²/16

Use the invariant transformations:
4·(a²+b²+c²+d²) ≥
≥ a²+b²+c²+d²+
+2ab+2ac+2ad+
+2bc+2bd+2cd

3·(a²+b²+c²+d²) ≥
≥ 2ab+2ac+2ad+
+2bc+2bd+2cd

(a−b)²+(a−c)²+(a−d)²+
+(b−c)²+(b−d)²+(c−d)² ≥ 0
which is obvious, and all transformations are reversible.
Therefore, the proof is to derive the required inequality by reversed transformations from the obviously correct last inequality.

Proof C
Given N positive real numbers X_i (i∈[1,N]).
Notice that
[Σ_{1≤ i ≤N}X_i]² =
= Σ_{1≤ i ≤N}X_i² +
+ Σ_{1≤ i < j ≤N}2X_iX_j

Let's start from the obvious
Σ_{1≤ i < j ≤N}(X_i−X_j)² ≥ 0
From this follows:
(N−1)·Σ_{1≤ i ≤N}X_i² −
− Σ_{1≤ i < j ≤N}2X_iX_j ≥ 0
Move products of different numbers to the right side of the inequality and add the sum of the squares of all numbers to both sides.
N·Σ_{1≤ i ≤N}X_i² ≥
≥ Σ_{1≤ i ≤N}X_i² +
+ Σ_{1≤ i < j ≤N}2X_iX_j
Transform the right side into a square of the sum of all numbers.
N·Σ_{1≤ i ≤N}X_i² ≥
≥ [Σ_{1≤ i ≤N}X_i]²


Summary

Harmonic mean (HM)
N / [Σ_i∈[1,N]1/X_i]
is less or equal to
geometric mean (GM)
[Π_i∈[1,N]X_i]^1/N
which is less or equal to
arithmetic mean (AM)
[Σ_i∈[1,N]X_i] / N
which is less or equal to
quadratic mean (QM)
sqrt{[Σ_i∈[1,N]X_i²] / N}

HM ≤ GM ≤ AM ≤ QM