An Inequality for Grade 8

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

Assume $n$ positive numbers $(n \gt 0)$ $x_{k}$ add up to $\displaystyle\frac{1}{2}$:

$\displaystyle x_{1} + ... + x_{n} = \frac{1}{2}.$

Prove that

$\displaystyle\frac{1-x_1}{1+x_1}\cdot\frac{1-x_2}{1+x_2}\cdot\ldots\cdot\frac{1-x_n}{1+x_n}\ge\frac{1}{3}.$

Proof

Assume $n$ positive numbers $(n \gt 0)$ $x_{k}$ add up to $\displaystyle\frac{1}{2}$:

$\displaystyle x_{1} + ... + x_{n} = \frac{1}{2}.$

Prove that

$\displaystyle\frac{1-x_1}{1+x_1}\cdot\frac{1-x_2}{1+x_2}\cdot\ldots\cdot\frac{1-x_n}{1+x_n}\ge\frac{1}{3}.$

Proof

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

Lemma

For positive $a$ and $b,$

$\displaystyle\frac{1-a}{1+a}\frac{1-b}{1+b}\ge\frac{1-a-b}{1+a+b}.$

This is shown by direct verification. Introduce function

$\displaystyle f(x)=\frac{1-x}{1+x}.$

Lemma asserts that $f(x) f(y)\ge f(x + y).$ For the inductive step, we want to show that if $\displaystyle x_{1} + \ldots + x_{n} = \frac{1}{2}$ implies

$\displaystyle f(x_{1}) f(x_{2}) \ldots f(x_{n}) \ge \frac{1}{3},$

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

$\displaystyle f(x_{1}) f(x_{2}) \ldots f(x_{k-1})f(x_{k}) \ge f(x_{1}) f(x_{2}) \ldots f(x_{k-1} + x_{k})\ge \frac{1}{3}$

because, by the assumption, $\displaystyle x_{1} + \ldots + (x_{k-1} + x_{k}) = \frac{1}{2}$. 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 $\displaystyle\frac{1}{2}?$

Well, the only time that condition was used happened at the very beginning: for $n = 1.$ If $\displaystyle x_{1} = \frac{1}{2},$ then indeed

$\displaystyle f(\frac{1}{2}) = (1 - \frac{1}{2}) / ( 1 + \frac{1}{2}) = \frac{1}{3} \ge \frac{1}{3}.$

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

$\displaystyle\frac{1-x_1}{1+x_1}\cdot\frac{1-x_2}{1+x_2}\cdot\ldots\cdot\frac{1-x_n}{1+x_n}\ge\frac{1}{2}.$

Starting with $\displaystyle \frac{1}{2}$ leads to $\displaystyle \frac{1}{3};$ starting with $\displaystyle \frac{1}{3}$ leads to $\displaystyle \frac{1}{2}!$ Interesting, is it not? Are the numbers $\displaystyle \frac{1}{2}$ and $\displaystyle \frac{1}{3}$ 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 $x_{1} + \ldots + x_{n} = a$ implies $f(x_{1}) \ldots f(x_{n}) \ge b,$ then $x_{1} + \ldots + x_{n} = b$ implies $f(x_{1}) \ldots f(x_{n}) \ge a.$ (Just in passing, functions for which $f(f(x))=x$ are called involutions.)

Reference

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

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