PROOF 192
Prove.
The arithmetic mean is at least the geometric mean
On the non-negative reals.
A way to read this proof
First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.
Theorem T1
For real numbers a, b ≥ 0, (a + b)/2 ≥ √(ab)
Proof.
(1)a ≥ 0 and b ≥ 0, so √a and √b exist and (√a)² = a, (√b)² = bassumption, A2
(2)(√a − √b)² ≥ 0A1
(3)(√a − √b)² = (√a)² − 2·√a·√b + (√b)²A3
(4) = a − 2√(ab) + b(1), (3), A6, E
(5)a − 2√(ab) + b ≥ 0(2), (4)
(6)Adding 2√(ab) to both sides gives a + b ≥ 2√(ab)(5), A4
(7)Dividing both sides by 2 gives (a + b)/2 ≥ √(ab)(6), A5
∎
Theorem T2
(a + b)/2 = √(ab) holds if and only if a = b.
Proof.
(1)Suppose equality holds. Multiplying both sides by 2 gives a + b = 2√(ab)assumption, E
(2)Subtracting 2√(ab) from both sides gives a − 2√(ab) + b = 0(1), E
(3)(√a − √b)² = a − 2√(ab) + bA2, A3, A6
(4)(√a − √b)² = 0(2), (3)
(5)√a − √b = 0, that is, √a = √b(4), A7, E
(6)a = (√a)² = (√b)² = b(5), A2, E
(7)Conversely, suppose a = b. Then √a = √bassumption, E
(8)(√a − √b)² = 0² = 0, so a − 2√(ab) + b = 0(7), (3), N, E
(9)Adding 2√(ab) and dividing by 2 gives (a + b)/2 = √(ab)(8), E
∎
Corollary C
For real numbers a, b, c, d ≥ 0, (a + b + c + d)/4 ≥ ⁴√(abcd)
Proof.
(1)⁴√x denotes √√x.convention
(2)(a + b)/2 ≥ √(ab), (c + d)/2 ≥ √(cd)T1
(3)Adding side by side and dividing by 2 gives (a+b+c+d)/4 ≥ (√(ab) + √(cd))/2(2), A4, A5, N
(4)Applying the theorem again to the two non-negative numbers √(ab), √(cd) gives (√(ab) + √(cd))/2 ≥ √(√(ab)·√(cd))A2, T1
(5)√(√(ab)·√(cd)) = √√(abcd) = ⁴√(abcd)A6, (1), E
(6)(a + b + c + d)/4 ≥ ⁴√(abcd)(3), (4), (5), A8
(7)Repeating the same step extends this to n = 2ᵏ numbers. The inequality for general n is not proved here and is left as a statement.(6)
∎
Remark
Can A1 be removed from this proof?
The whole inequality stands on a single fact: the square of a real number is non-negative.