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

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?
