## 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, prove

- a
^{4}+ b^{4}≥ 2a^{2}b^{2}- a
^{4}+ b^{4}+c^{4}+ d^{4}≥ 4abcd1. aUsing equation (1) replacing a with a^{4}+ b^{4}≥ 2a^{2}b^{2}^{4}and b with b^{4}, then (a^{4}+ b^{4})/2 ≥ √(a^{4}b^{4}) ≥ a^{2}b^{2}

∴ a^{4}+ b^{4}≥ 2a^{2}b^{2}.... (2)

2. aUsing equation 2 a^{4}+ b^{4}+c^{4}+ d^{4}≥ 4abcd^{4}+ b^{4}+c^{4}+ d^{4}≥ 2a^{2}b^{2}+ 2c^{2}d^{2}Using equation (1) replacing a with 2a^{2}b^{2}and b with 2c^{2}d^{2}, then (2a^{2}b^{2}+ 2c^{2}d^{2})/2 ≥ √(2a^{2}b^{2}.2c^{2}d^{2}) i.e. 2a^{2}b^{2}+ 2c^{2}d^{2}≥ 4abcd so a^{4}+ b^{4}+c^{4}+ d^{4}≥ 2a^{2}b^{2}+ 2c^{2}d^{2}≥ 4abcd∴ a^{4}+ b^{4}+c^{4}+ d^{4}≥ 4abcd

### Example 8.

Show if a, b, c are real numbers then a^{2}+ b^{2}+ c^{2}- bc - ca - ab cannot be negative. From (1) we have (a + b)/2 ≥ √(ab) or (a + b) ≥ 2√(ab) So replacing a with a^{2}and b with b^{2}we have a^{2}+ b^{2}≥ 2√(a^{2}b^{2}) ie a^{2}+ b^{2}≥ 2ab Similarly b^{2}+ c^{2}≥ 2bc c^{2}+ a^{2}≥ 2ca Add the 3 equations together 2(a^{2}+ b^{2}+c^{2}) ≥ 2(ab + bc + ca)∴ (a^{2}+ b^{2}+c^{2}) ≥ ab + bc + ca

### Exercise12 .

Show when a,b are positive numbersa + 1/a ≥ 2In 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) ≥ 8abcIn 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 x^{3}+ y^{3}> x^{2}y + xy^{2}---(a) if (x + Y) > 0.Note(a) should really be ≥ and not > because if x=y then (a) reduces to 2x^{3}> 2x^{3}which clearly not the case. Anyway progressing on the assumption x ⧧ y. Consider (x - y)^{2}(x + y) = (x + y)(x^{2}- 2xy + y^{2}) = x^{3}- 2x^{2}y + x^{2}y + xy^{2}- 2xy^{2}+y^{3}) = x^{3}- x^{2}y - xy^{2}+y^{3}Now (x - y)^{2}(x + y) is clearly greater than zero if (x + y) > 0 and x ⧧ y. ∴ x^{3}+ y^{3}- x^{2}y - xy^{2}> 0∴ x^{3}+ y^{3}> x^{2}y + xy^{2}