Here is a complete solution proposal for the case $n\geq 4$.
Sorry for the bad english.
The set of $(x_1,...,x_n)$ with non negative coefficients and sum $1$ is compact and the function $f$ considered is continuous, so we can fix $(a_1,...,a_n)$ a point where $f$ reaches its minimum.
Among all these points, we can consider one that has a maximal number of zeros in the list $(a_1,...,a_n)$.
Let us reason by contradiction and assume that this maximal number is less than or equal to $n-3$ and let $(a_1,...,a_n)$ be a associated point (which thus has at least three positive coefficients).
By circular permutation, we can assume without loss of generality that $a_1>0$ and $a_j>0$ for some $j\in [3,n-1]$.
We set $S = a_1+a_j$ and $g$ the function defined on $[0,S]$ by
$g(x) = f(x,a_2,...,a_{j-1},S-x,a_{j+1},....,a_n)$.
There exists a constant $C$ independent of $x$ such that
$g(x) = \dfrac{x+a_2}{1+a_2 x}+\dfrac{x+a_n}{1+a_n x} + \dfrac{a_{j-1}+S-x}{1+a_{j-1}(S-x)} + \dfrac{a_{j+1}+S-x}{1+a_{j+1}(S-x)}+C.$
Note that $j\in [3,n-1]$ allows $x$ and $S-x$ not to mix in the fractions.
For any $a\in [0,1]$, we set $h_a(x) = \dfrac{a+x}{1+ax}$ and we have $h_a''(x) = -\dfrac{2a(1-a^2)}{(1+ax)^3}\leq 0$, so each function $h_a$ is concave on $[0,S]$.
By symmetry, each function $x\mapsto h_a(S-x)$ is also concave on $[0,S]$.
By summing concave and continuous functions, the function $g$ is concave and continuous and therefore, $\min g_{[0,S]} = \min (g(0),g(S))$.
Thus, we have $\min f = f(a_1,...,a_n) = g(a_1) \geq \min (g(0),g(S)) = \min (f(0,a_2,...,a_{j-1},a_1+a_j,...,a_n), f(a_1+a_j,a_2,...,a_{j-1},0,....,a_n))$,
so $f$ has a minimum at $(0,a_2,...,a_{j-1},a_1+a_j,...,a_n)$ or $(a_1+a_j,a_2,...,a_{j-1},0,....,a_n)$, which contradicts the assumed maximality of the number of zeros in $(a_1,...,a_n)$ (one of the two strictly positive terms $a_1$ or $a_j$ could have been replaced by 0 while maintaining the minimality property).
After this proof by contradiction, we know that $f$ has a minimum at some point $(a_1,...,a_n)$ with at least $n-2$ zeros.
We have $f(1,0,...,0)=2$ and $f(1/2,1/2,0,....,0) = \dfrac{9}{5}<2$, so we can affirm that the maximum number of zeros that we were interested in from the beginning is $n-2$.
Therefore, there exists a point $(a_1,...,a_n)$ such that $a_1>0$, $a_j>0$ for some $j\in [2,n]$, and $a_i=0$ for all $i\neq 1,j$ that satisfies $\min f = f(a_1,...,a_n)$.
If $j\in [3,n-1]$, by a completely similar reasoning (using the concavity argument), we have
$\min f \geq \min f(a_1+a_j,a_2,..,0,a_{j+1},...,a_n) , f(0,...,a_1+a_j,...,a_n) )=f(1,0,...,0,0,...,0)=2$
which is absurd.
Therefore, by circular permutation, we can find a point $(a_1,a_2,0,...,0)$ such that $\min f = f(a_1,a_2,0,...,0)$. To conclude, we only need to minimize $x\mapsto f(x,1-x,0,...,0)$, which is achieved at $x=1/2$, showing that $\min f\geq f(1/2,1/2,0,...,0)=\dfrac{9}{5}$.
Best Answer
By AM-GM $$1-x_i=\sum_{k\neq i}x_k\geq(n-1)\sqrt[n-1]{\prod_{k\neq i}x_k}.$$ Thus, $$\prod_{i=1}^n(1-x_i)\geq(n-1)^n\prod_{i=1}^n x_i$$ and we are done!