This is slightly too long to be comment, though this is not an answer:
Edit: I am providing a few revisions, but I still have not found a proof.
My previous comment was not quite correct. I "rearranged" the inequality by raising both sides of the inequality to certain powers, and I did not take into account that we do not know a priori whether any of the terms are larger or smaller than $1$. Depending on the size of each term (in particular, whether it is smaller or larger than $1$), exponentiating both sides could potentially reverse the inequality. I have returned the question to its original form as suggested by the OP, and I have added one more observation.
Let me first say that I have been working on this question for a bit, and though I have not yet resolved it, I have been having fun trying!
Now, to emphasize the dependence on $n$, let's set
$$
\alpha_n = \sum_{i=1}^n a_i \qquad \beta_n = \sum_{i=1}^n b_i, \qquad \sigma_n = \sum_{i=1}^n c_i,
$$
where $c_i = \sqrt{a_i b_i}$. Further, let's put
$$
A_n = \prod_{i=1}^n (a_i)^{a_i}, \qquad B_n = \prod_{i=1}^n (b_i)^{b_i}, \qquad S_n = \prod_{i=1}^n (c_i)^{c_i}.
$$
Our goal is to show:
\begin{equation}
(1) \hspace{1in} \left(\frac{A_n}{(\alpha_n)^{\alpha_n}}\right)^{\frac{1}{\alpha_n}} \cdot \left(\frac{B_n}{(\beta_n)^{\beta_n}} \right)^{\frac{1}{\beta_n}} \leq \left(\frac{S_n}{(\sigma_n)^{\sigma_n}}\right)^{\frac{2\sigma_n}{\alpha_n \beta_n}}
\end{equation}
A few pedestrian observations:
If $a_i = b_i$ for $i = 1, \dots , n$ (which forces $c_i = a_i = b_i$), then $A_n = B_n = S_n$, we also have $\alpha_n = \beta_n = \sigma_n$, and (1) holds in this case.
Note that $2c_i \leq a_i + b_i$ as $2xy \leq x^2 + y^2$ for all real numbers $x, y$. Hence, $2\sigma_n \leq \alpha_n + \beta_n$. Furthermore, Cauchy-Schwarz gives $\sigma_n^2 \leq \alpha_n \beta_n$. Both of these observations imply that $(\sigma_n + 1)^2 \leq (\alpha_n + 1)(\beta_n + 1)$.
I would imagine that with enough creativity, one may find a proof of the inequality involving convexity or a simple application of the AM-GM inequality (which I suppose is much the same thing!).
I have been unable to prove the inequality even in the case $n = 2$, when I assume $\alpha_n = \beta_n = 1$. I am not hopeful for a proof of the general case.
Since $e^x\geq 1+x$ for all $x\in \mathbb{R}$, by letting $x=-\log u$ we have that for any $u >0$, $\log u\geq 1-\frac{1}{u}$, and so $u\log u>u-1$ for any $u>0$. Combining these two inequalities, it follows that $$\sqrt{x^x y^y}= \exp\left(\frac{x\log x}{2}+\frac{y\log y}{2}\right)\geq \exp\left(\frac{x+y}{2}-1\right)\geq \frac{x+y}{2}.$$ Hence $$\frac{x^x}{x+y}+\frac{y^y}{y+z}+\frac{z^z}{z+x}\geq \frac{1}{2} \left(\sqrt{\frac{x^x}{y^y}}+\sqrt{\frac{y^y}{z^z}}+\sqrt{\frac{z^z}{x^x}}\right),$$ and the result follows from the AM-GM inequality.
Remark: By combining $e^x\geq 1+x$ for $x\in\mathbb{R}$ and $u\log u\geq u-1$ for $u>0$ as above, we obtain the following "reverse exponential AM-GM inequality": $$\sqrt[n]{x_{1}^{x_{1}}x_{2}^{x_{2}}\cdots x_{n}^{x_{n}}}\geq \frac{x_1+x_2+\cdots+x_n}{n}.$$
Best Answer
For any $x_1,\ldots,x_n \in [0,\frac12]$, define $$f(x_1,\ldots,x_n) = \frac{\frac1n\sum_{k=1}^n x_k}{1 - \frac1n\sum_{k=1}^n x_k}$$ The inequality at hand can be rewritten as $$f(x_1,\ldots,x_n) \ge \prod_{k=1}^n f(x_k)^{\frac1n}\tag{*1}$$ For $n \ge 2$, let $\mathcal{S}_n$ be statement $(*1)$ is true for any $(x_1,\ldots,x_n) \in [0,\frac12]$.
$\mathcal{S}_2$ is true.
For any $x,y \in [0,\frac12]$, we have $$f(x,y)^2 - f(x)f(y) = \frac{(1-x-y)(x-y)^2}{(1-x)(1-y)(2-x-y)^2} \ge 0$$
Assume $\mathcal{S}_n$ is true.
For any $x_1,x_2,\ldots, , x_{2n} \in [0,\frac12]$, let $z_k = \frac12(x_{2k-1} + x_{2k})$ for $k = 1,\ldots, n$.
Since $\frac{1}{2n}\sum_{k=1}^{2n} x_n = \frac{1}{n}\sum_{k=1}^n z_k$, we have $$\begin{align} f(x_1,\ldots,x_{2n}) = & f(z_1,\ldots,z_n)\\ \stackrel{\mathcal{S}_n}{\ge} & \prod_{k=1}^n f(z_k)^{\frac1n} = \prod_{k=1}^n f(x_{2k-1},x_{2k})^{\frac1n}\\ \stackrel{\mathcal{S}_2}{\ge} & \prod_{k=1}^n \left( f(x_{2k-1})^{\frac12}f(x_{2k})^{\frac12} \right)^{\frac1n} = \prod_{k=1}^{2n} f(x_k)^{\frac{1}{2n}}\end{align}$$ This means $\mathcal{S}_n \implies \mathcal{S}_{2n}$.
Assume $\mathcal{S}_{n+1}$ is true for some $n \ge 2$.
For any $x_1,x_2,\ldots, , x_{n} \in [0,\frac12]$, let $x_{n+1} = \bar{x} \stackrel{def}{=} \frac1n\sum_{k=1}^n x_k$.
If $\bar{x} = 0$, then all $x_k = 0$ and $(*1)$ is satisfied trivially.
Assume $\bar{x} \ne 0$. Notice $\displaystyle\;\frac{1}{n+1}\sum_{k=1}^{n+1} x_k = \bar{x} = \frac{1}{n}\sum_{k=1}^n x_k$, we find $$f(x_1,\ldots,x_n) = f(\bar{x}) = f(x_1,\ldots,x_{n+1}) \stackrel{\mathcal{S}_{n+1}}{\ge} \left(\prod_{k=1}^{n+1} f(x_k)\right)^{\frac{1}{n+1}} = \left(f(\bar{x})\prod_{k=1}^n f(x_k)\right)^{\frac{1}{n+1}}$$ Rising both side to power $\frac{n+1}{n}$ and remove a common factor $f(\bar{x})^{\frac1n}$, we find $(*1)$ is satisfied again. This means $\mathcal{S}_{n+1} \implies \mathcal{S}_n$.
By Forward-Backward induction, $\mathcal{S}_n$ is true for all $n \ge 2$.
Update
Look at this question again, I notice I have overlooked the part "the equality holds if and only if $x_1 = \cdots = x_n$". The modification of above proof to cover that is relatively straight forward. I'll leave that as an exercise.