I was hoping the generalize this result: How to prove $x^{x/(1+x)}/(1+x)\geq1/2$
I believe that the following inequality holds:
$\frac{\Pi_i x_i ^\frac{x_i}{1+\sum_i x_i}}{1+\sum_i x_i} \geq \frac{1}{N+1}$
Where $x_i>0$ and $N$ is the number of elements.
Edit Here is my attempt (trying to use the method from linked question)
By bernoulli’s inequality
$x_i ^\frac{x_i}{1+\sum_i x_i}= \frac{1}{1+\left( \frac{1}{x_i} -1\right)^\frac{x_i}{1+\sum_i x_i} } \geq \frac{1}{1+\left( \frac{1}{x_i} -1\right) \left( \frac{x_i}{1+\sum_i x_i}\right) } =\frac{1+\sum_i x_i}{2-x_i +\sum_i x_i}$
thus
$\frac{\Pi_i x_i ^\frac{x_i}{1+\sum_i x_i}}{1+\sum_i x_i} \geq \frac{\left({1+\sum_i x_i}\right)^{N-1}} { \Pi_i \left(2-x_i +\sum_i x_i\right)}$
Then by the AM-GM inequality
$\Pi_i \left(2-x_i +\sum_i x_i\right) \leq \left(\frac{\sum_i \left( 2-x_i +\sum_i x_i\right)}{N}\right)^N = \left(\frac{2N +(N-1)\sum_i x_i }{N}\right)^N = \left( 2 +\left(1-\frac{1}{N}\right)\sum_i x_i \right)^N$
thus
$\frac{\Pi_i x_i ^\frac{x_i}{1+\sum_i x_i}}{1+\sum_i x_i} \geq \frac{\left({1+\sum_i x_i}\right)^{N-1}} { \Pi_i \left(2-x_i +\sum_i x_i\right)} \geq \frac{\left({1+\sum_i x_i}\right)^{N-1}} { \left( 2 +\left(1-\frac{1}{N}\right)\sum_i x_i \right)^N} $
Let's decompose the denominator of the $LHS$ into $n+1$ terms: $$\begin{align} \frac{\prod_i x_k ^\frac{x_k}{1+\sum_i x_i}}{1+\sum_i x_i} &= \frac{\prod_i x_k ^\frac{x_k}{1+\sum_i x_i}}{(1+\sum_i x_i)^{\frac{1}{1+\sum_i x_i}}\cdot \Pi_k\left((1+\sum_i x_i)^{\frac{x_k}{1+\sum_i x_i}}\right)} \\ &=\left(\frac{1}{1+\sum_{i=1}^nx_i} \right)^{\frac{1}{1+\sum_{i=1}^nx_i}} \prod_{k=1}^n \left(\frac{x_k}{1+\sum_{i=1}^nx_i} \right)^{\frac{x_k}{1+\sum_{i=1}^nx_i}} \end{align}$$
Denote $y_0 = \frac{1}{1+\sum_{i=1}^nx_i}$ and $y_k = \frac{x_k}{1+\sum_{i=1}^nx_i}$ for $k = 1,...,n$, then the problem is equivalent to $$\begin{align}&\prod_{k=0}^n y_k^{y_k} \ge\frac{1}{n+1} \\ &\iff \sum_{k=0}^n y_k \ln(y_k) \ge \ln \left( \frac{1}{n+1} \right)\\ & \iff \frac{1}{n+1} \sum_{k=0}^n y_k \ln(y_k) \ge \frac{1}{n+1} \cdot\ln \left( \frac{1}{n+1} \right)\tag{1}\end{align}$$ given $y_k> 0$ and $\sum_{k=0}^ny_k=1$.
To prove $(1)$, it suffices to apply the Jensen's inequality to the convex function $f(t)= t\ln(t)$ (indeed, $f(t)$ is convex as $f''(t) = 1/t >0$ for $t>0$): $$ \frac{1}{n+1} \sum_{k=0}^n y_k \ln(y_k) \ge \frac{\sum_{k=0}^ny_k}{n+1}\cdot \ln \left(\frac{\sum_{k=0}^ny_k}{n+1}\right) = \frac{1}{n+1} \cdot\ln \left( \frac{1}{n+1} \right)$$
The equality occurs if and only if all $y_k$ are equal, or $x_1=x_2=...=x_n = 1$.