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The question is this

The sum of positive numbers $x_1,x_2,x_3,\dotsc,x_n$ is $\frac{1}{2}$. Prove that $$\frac{1-x_1}{1+x_1}\cdot\frac{1-x_2}{1+x_2}\cdot\frac{1-x_3}{1+x_3}\cdots\frac{1-x_n}{1+x_n}\geq\frac{1}{3}.$$

My process was something like this:

Using Weirstrass' Inequality in the numerators, we get $$\prod_{i=1}^{n}(1-x_i)>1-\sum_{i=1}^{n}(x_i)=1-\frac12=\frac12.$$ So now we need to prove that $$\prod_{i=1}^{n}(1+x_i)<\frac32.$$

But then, again using Weirstrass' Inequality, we get $$\prod_{i=1}^{n}(1+x_i)>1+\sum_{i=1}^{n}(x_i)=\frac32.$$

So now I'm stuck. Please help me solve this problem.

Note: This question if from Mathematical Circles (Russian Experience) so I don't think that this question is supposed to be wrong

Bart Michels
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user117913
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  • From my experience, using strict inequalities when the question involves non-strict inequalities usually leads nowhere. – MT_ Jan 31 '14 at 16:22
  • @user92774 So how do you suggest that I move forward? I think that the equality holds iff there is only one term,i.e, $x_i$ . For that case, equality holds even in my process. – user117913 Jan 31 '14 at 16:31
  • Yes, that seems to be the case. – MT_ Jan 31 '14 at 16:33
  • Using AM-GM one can guess that $S \geq (\frac{2n-1}{2n+1})^n \geq \frac{1}{3}$ but I wouldn't know how to formalize that argument. – MT_ Jan 31 '14 at 16:41
  • @user92774 I had tried to do what you guessed and obtained the numerator, but I was unable to get the denominator. – user117913 Jan 31 '14 at 16:49
  • Induction seems to be the way to go here, since both numerator and denominator, if treated separately, will have the same direction of inequality. – MT_ Jan 31 '14 at 16:54

2 Answers2

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We will prove it using induction over $n$.

  1. It is seen that for $n=1$ it follows that $x_{1} = \frac{1}{2}$. Then $$ \frac{1-x_{1}}{1+x_{1}} = \frac{1}{3}. $$

  2. Assume it holds for $n=k$, then for $n = k+1$, we have $\sum_{i=1}^{k+1} x_{i} = \frac{1}{2}$. Define the element $y_{i} = x_{i}$ for $i=1, \ldots,n-1$ and $y_{n} = x_{n}+x_{n+1}$. Then still $\sum_{i=1}^{k} y_{i} = \frac{1}{2}$ and by assumption we have: $$ \frac{1-y_{1}}{1+y_{1}} \cdot \frac{1-y_{2}}{1+y_{2}} \cdots \frac{1-y_{k}}{1+y_{k}} \geq \frac{1}{3}. $$ It is seen that $$ \frac{1-x_{k}}{1+x_{k}} \cdot \frac{1-x_{k+1}}{1+x_{k+1}} = \frac{1 - x_{k} - x_{k+1} + x_{k} x_{k+1}}{1+ x_{k} + x_{k+1} + x_{k} x_{k+1}} > \frac{1 - x_{k} -x_{k+1}}{1+x_{k}+x_{k+1}} = \frac{1-y_{k}}{1+y_{k}}. $$ Combining, we have: $$ \frac{1-x_{1}}{1+x_{1}} \frac{1-x_{2}}{1+x_{2}} \cdots \frac{1-x_{k}}{1+x_{k}} \cdot \frac{1-x_{k+1}}{1+x_{k+1}} \geq \frac{1-y_{1}}{1+y_{1}} \cdot \frac{1-y_{2}}{1+y_{2}} \cdots \frac{1-y_{k}}{1+y_{k}} \geq \frac{1}{3}. $$

Vincent
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    Comparing your proof with mine, I see that I tried to prove $\frac{1-x_k}{1+x_k}\frac{1-x_{k+1}}{1+x_{k+1}}>\frac{1-y_k}{1+y_k}$ by taking the numerator and denominator separately. I suppose that is where I made my mistake. – user117913 Jan 31 '14 at 17:14
  • It is. Remains one question for me: given a number $n$ what positive (or non-negative) $x_{i}$ do I need to choose such that the product is maximal. It seems to me that this is maximal if we choose $x_{i} = \frac{1}{2 n}$ all equal. The maximum would then be $\left( \frac{2n - 1}{2n + 1} \right)^{n} \to \frac{1}{e}$ if $n \to \infty$. – Vincent Feb 03 '14 at 11:49
  • @Vincent You can prove the maximum by smoothing. Show that for $ a < c < b \leq 1/2$, we have $ (1-a)/(1+b)(1-b)/(1+b) \leq (1-c)/(1+c) ( 1-a-b+c)/(1+a+b-c )$. So we replace the extreme elements with $ 1/2n, a+b - 1/2n$ and repeat that. – Calvin Lin Jan 08 '22 at 00:59
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As you say in the comment, equality holds if there is just one positive $x_i$.

We need to prove that adding another positive $x_j$ (and thereby reducing $x_i$) will increase the ratio.

Let $x_j = x$ and $x_i = \frac{1}{2} - x$.

Then the ratio is $\frac{(1 - x)(\frac{1}{2} +x)}{(1+ x)(\frac{3}{2} - x)} = \frac{\frac{1}{2} + x(\frac{1}{2} - x)}{\frac{3}{2} + x(\frac{1}{2} - x)} \ge \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{3}$.

Can you extend that argument ?by induction on the number of terms?

oks
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