# If $\sum x_k=\frac12$, then $\prod\frac{1-x_k}{1+x_k}\geq\frac13$

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 $$\prod_{i=1}^{n}(1+x_i)>1+\sum_{i=1}^{n}(x_i)=\frac32.$$

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

• From my experience, using strict inequalities when the question involves non-strict inequalities usually leads nowhere. – MCT 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. – MCT 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. – MCT 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

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: $$\frac{1-x_{1}}{1+x_{1}} \frac{1-x_{2}}{1+x_{2}} \cdots \frac{1-x_{k}}{1+x_{k}} \cdot \frac{1-x_{k+1}}{1+x_{k+1}} \geq \frac{1-y_{1}}{1+y_{1}} \cdot \frac{1-y_{2}}{1+y_{2}} \cdots \frac{1-y_{k}}{1+y_{k}} \geq \frac{1}{3}.$$

• 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 3 '14 at 11:49

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?

• I did try it, but it proved too difficult (at least for me). – user117913 Jan 31 '14 at 16:52
• @ oks: Oops, saw your hint, only after I copied down mine, took me some time. – Vincent Jan 31 '14 at 17:03
• @Vincent nice, thank you! – oks Jan 31 '14 at 17:04