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

We will prove it using induction over $n$.


*

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

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