Is this function always bounded below? Is there any global constant $C$ such that $$C<\frac{\sum_{i=1}^{n}x_{i}\log x_{i}-x_{i}+(1-x_{i})\log(1-x_{i})}{\sum_{i=1}^{n}x_{i}}+\log(\sum_{i=1}^{n}x_{i})-\log(\sum_{i=1}^{n}x_{i}^{2})$$ for all vectors $x\in\mathbb{R}^n$ such that $0<x_i<1$ for all $i$?  The main sticking point seems to be that the last term is not convex, so i can't just fix $\sum_i x_i $ and apply symmetry.
 A: Your expression is not bounded from below by an absolute constant (independent of $n$). To see this, consider the example $x_1=\dots=x_{n-1}=\epsilon$ and $x_n=1-\epsilon$. For this example, the left hand side equals
$$ n\cdot\frac{\epsilon\log\epsilon+(1-\epsilon)\log(1-\epsilon)}{1+(n-2)\epsilon}-1+\log\bigl(1+(n-2)\epsilon\bigr)-\log\bigl(1-2\epsilon+n\epsilon^2\bigr).$$
For $\epsilon:=1/n$, the first term asymptotically $-(\log n)/2$, while the other three terms are bounded, so the expression tends to negative infinity.
A: Now I think that no, even if we remove the negative summands $-x_i+(1-x_i)\log(1-x_i)$. Note that our inequality becomes homogeneous, so we may forget that $x_i$ are less than 1. Choose $x_i=1+t_i$ so that $\sum t_i=0$, then the inequality rewrites as 
$$
\log\frac{\sum x_i^2}{\sum x_i}=\log\left(1+\frac{t_1^2+\dots+t_n^2}n\right)<-C+\frac1n\sum (1+t_i)\log(1+t_i).
$$
Imagine that $t_1=n^{3/4}$ and other $t$'s are equal to $-t_1/(n-1)$. Then LHS is large while RHS is bounded.
