An Inequality for Grade 8

The following problem was offered at the 54th Leningrad Mathematical Olympiad (1988) for grades 8-10 (the last three grades in the Russian system) students.

Assume n positive numbers (n > 0) xk add up to 1/2:

x1 + ... + xn = 1/2.

Prove that

(1 - x1)

(1 + x1)
(1 - x2)

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

(1 + xn)
1

3

Proof

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Copyright © 1996-2012 Alexander Bogomolny

Assume n positive numbers (n > 0) xk add up to 1/2:

x1 + ... + xn = 1/2.

Prove that

(1 - x1)

(1 + x1)
(1 - x2)

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

(1 + xn)
1

3

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 + ... + xn = 1/2 implies

f(x1) f(x2) ... f(xn) ≥ 1/3,

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

 f(x1) f(x2) ... f(xk-1)f(xk) ≥ f(x1) f(x2) ... f(xk-1 + xk)
  1/3,

because, by the assumption, x1 + ... + (xk-1 + xk) = 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

f(1/2) = (1 - 1/2) / ( 1 + 1/2) = 1/31/3.

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 + ... + xn = a implies f(x1) ... f(xn) ≥ b, then x1 + ... + xn = b implies f(x1) ... f(xn) ≥ a.

Reference

  1. D. Fomin, A. Kirichenko, Leningrad Mathematical Olympiads 1987-1991, MathPro Press, 1994

Inequalities to prove:

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Copyright © 1996-2012 Alexander Bogomolny

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