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.



Is this function always bounded below?
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.


