Inequality of Means
Arithmetic ≥ Mean Geometric
Consider positive real numbers a, b Arithmetic Mean = (a + b)/2 Geometric Mean = √(ab) Show (a - b)/2 ≥ √(ab) Now (√a + √b)2 ≥ 0 So a + b -2√(ab) ≥ 0 (a + b)/2 ≥ √(ab) .......(1)
Examples based upon Arithmetic ≥ Mean Geometric
The following exercises and examples are taken from Pure Mathematics For Advanced Level by Bunday and Mulholland, pages 14 and 15.
Example:7
If a, b, c, d are any real numbers, prove1. a4 + b4 ≥ 2a2b2 Using equation (1) replacing a with a4 and b with b4, then (a4 + b4)/2 ≥ √(a4b4) ≥ a2b2
- a4 + b4 ≥ 2a2b2
- a4 + b4 +c4 + d4 ≥ 4abcd
∴ a4 + b4 ≥ 2a2b2 .... (2)
2. a4 + b4 +c4 + d4 ≥ 4abcd Using equation 2 a4 + b4 +c4 + d4 ≥ 2a2b2 + 2c2d2 Using equation (1) replacing a with 2a2b2 and b with 2c2d2, then (2a2b2 + 2c2d2)/2 ≥ √(2a2b2.2c2d2) i.e. 2a2b2 + 2c2d2 ≥ 4abcd so a4 + b4 +c4 + d4 ≥ 2a2b2 + 2c2d2 ≥ 4abcd ∴ a4 + b4 +c4 + d4 ≥ 4abcd
Example 8.
Show if a, b, c are real numbers then a2 + b2 + c2 - bc - ca - ab cannot be negative. From (1) we have (a + b)/2 ≥ √(ab) or (a + b) ≥ 2√(ab) So replacing a with a2 and b with b2 we have a2 + b2 ≥ 2√(a2b2) ie a2 + b2 ≥ 2ab Similarly b2 + c2 ≥ 2bc c2 + a2 ≥ 2ca Add the 3 equations together 2(a2 + b2 +c2) ≥ 2(ab + bc + ca) ∴ (a2 + b2 +c2) ≥ ab + bc + ca
Exercise12 .
Show when a,b are positive numbers a + 1/a ≥ 2 In 1 above it has been shown (a + b)/2 ≥ √(ab) .......(1) Substitute 1/a for b and rearrange gives a + 1/a ≥ 2√(a/a) ≥ 2 and show (a + b)(1/a + 1/b) ≥ 4 (a + b)(1/a + 1/b) = (a + b)((a + b)/ab) = (a + b)2/ab from 1 (a + b)2 ≥ (2 √(ab))2 ≥ 4ab ∴ (a + b)2/ab ≥ 4 ∴ (a + b)(1/a + 1/b) ≥ 4
Exercise13 .
Show when a,b,c are positive numbers (a + b)(b + c)(c + a) ≥ 8abc In 1 above it has been shown (a + b) ≥ 2√(ab) .......(1) ∴ (b + c) ≥ 2√(bc) (c + a) ≥ 2√(ca) ∴ (a + b)(b + c)(c + a) ≥ 2√(ab).2√(bc).2√(ca) ≥ 8√(a.a.b.b.c.c) ∴ (a + b)(b + c)(c + a) ≥ 8abc
Exercise14 .
Show that x3 + y3 > x2y + xy2 ---(a) if (x + Y) > 0. Note (a) should really be ≥ and not > because if x=y then (a) reduces to 2x3 > 2x3 which clearly not the case. Anyway progressing on the assumption x ⧧ y. Consider (x - y)2(x + y) = (x + y)(x2 - 2xy + y2) = x3 - 2x2y + x2y + xy2 - 2xy2 +y3) = x3 - x2y - xy2 +y3 Now (x - y)2(x + y) is clearly greater than zero if (x + y) > 0 and x ⧧ y. ∴ x3 + y3 - x2y - xy2 > 0 ∴ x3 + y3 > x2y + xy2