Proof

The proof is by induction. For n = 1, the inequality is obvious. The inductive step is based on the following

Lemma

For positive a, b,

(1 - a) (1 + a) (1 - b) (1 + b) 1 - a - b 1 + a + b

This is shown by direct verification. Introduce function

f(x) = 1 - x 1 + x

Lemma asserts that f(x) f(y) ≥ f(x + y). For the inductive step, we want to show that if x₁ + ... + x_n = ¹/₂ implies

for n = k - 1, then the same also holds for n = k. Indeed, using Lemma

because, by the assumption, x₁ + ... + (x_k-1 + x_k) = ¹/₂. This completes the induction.

Now that we got through the proof, it is a good time to have a look back. Upon a cursory inspection, one question pops to mind: where did we use the condition that the given numbers add up to ¹/₂?

Well, the only time that condition was used happened at the very beginning: for n = 1. If x₁ = ¹/₂, then indeed

In the rest of the proof the ¹/₂ - requirement was used conditionally: if the sum of n terms was ¹/₂, we found (n - 1) term whose sum was still ¹/₂. The latter step remains valid if we replace ¹/₂ with any other number! So the final result hinges on the fact that f(¹/₂) = ¹/₃. This remark suggests a modification of the problem. For example, since f(¹/₃) = ¹/₂, we may claim that, for n numbers that add up to 1/3, the following inequality holds

(1 - x1) (1 + x1) (1 - x2) (1 + x2) ... (1 - xn) (1 + xn) 1 2 .

Starting with ¹/₂ leads to ¹/₃; starting with ¹/₃ leads to ¹/₂! Interesting, is it not? Are the numbers ¹/₂ and ¹/₃ form a special pair?

No, they are not. There are other such pairs; it is easy to find as many as you want. In fact it is practically impossible not to find some.

The reason is that function f has an interesting property of being its own inverse. More accurately, f(a) = b is equivalent to f(b) = a, as one can verify directly. So, in general, if x₁ + ... + x_n = a implies f(x₁) ... f(x_n) ≥ b, then x₁ + ... + x_n = b implies f(x₁) ... f(x_n) ≥ a.

Reference