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)
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
a4 + b4 ≥ 2a2b2
a4 + b4 +c4 + d4 ≥ 4abcd
1. a4 + b4 ≥ 2a2b2
Using equation (1) replacing a with a4 and b with b4, then
(a4 + b4)/2 ≥ √(a4b4)
≥ a2b2
∴ 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
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
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
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
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