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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.

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    $\begingroup$ What have you tried? $\endgroup$ – Dhanvi Sreenivasan May 22 at 10:37
  • $\begingroup$ @mtheorylord Do you mean $\prod_{i < j} (x_i - x_j)$ in the expression? $\endgroup$ – River Li May 22 at 11:55
  • $\begingroup$ @RiverLi, fixed. Dhanvi Sreenivasan, everything. $\endgroup$ – mtheorylord May 22 at 16:10
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    $\begingroup$ What have you tried? Where are you stuck? See How to ask a good question. $\endgroup$ – Saad May 25 at 1:48
  • $\begingroup$ @AlapanDas It is possible for the partial derivatives all to be negative. He is not saying that the function does not increase in any direction, but rather it does not increase in any direction along the positive $x_i$ axis, for any $x_i$. $\endgroup$ – paulinho May 25 at 3:16
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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?

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  • $\begingroup$ (+1) Your solution is impressive. $\endgroup$ – River Li May 28 at 4:17
  • $\begingroup$ That was clean. I've was stuck on it for a while, didn't think it had an easy solution. $\endgroup$ – mtheorylord May 28 at 6:09

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