TITLE: Change along some direction is positive
QUESTION [5 upvotes]: Let:
 $$f(x_1,\cdots,x_n) = \prod_{i}x_i(1-x_i) \prod_{i<j}|x_i-x_j|$$
Suppose all $x_i \in (0,1)$ are fixed and $\sum_{i}x_i < \frac{n}{2}$.
Show that there is some $i$ and a sufficiently small $\epsilon$ so that $x_i \mapsto x_i +\epsilon$ doesn't decrease the value of $f$.
That is to say, at least one of the partial derivatives of $f$ is non-negative.
After taking the derivative in each $x_i$ one gets a system of inequalities. I was able to prove the statement for $n=2,3$ this way through basically brute force. This doesn't generalize well though.

REPLY [3 votes]: It is actually not that difficult. Assuming that all $x_i$ are distinct and taking the logarithmic derivatives, we just need to show that at least one of the expressions 
$$
D_i=\frac 1{x_i}-\frac 1{1-x_i}+\sum_{j:j\ne i}\frac 1{x_i-x_j}
$$
is positive.
Now consider 
$$
\sum_i x_i(1-x_i)D_i=\sum_i(1-2x_i)+\sum_{i,j:i\ne j}\frac {x_i-x_i^2}{x_i-x_j}=\sigma+\Sigma\,.
$$
We have $\sigma=n-2\sum_ix_i>0$ by the assumption. Now we can use the antisymmetry of the denominator $x_i-x_j$ to write 
$$
2\Sigma=\sum_{i,j:i\ne j}\frac {(x_i-x_i^2)-(x_j-x_j^2)}{x_i-x_j}
\\
=
\sum_{i,j:i\ne j}(1-x_i-x_j)=n(n-1)-2(n-1)\sum_i x_i>0
$$
and we are done.
Is it just an exercise from some book or you really needed it for something?