# Math Genius: Change along some direction is positive

Let:
$$f(x_1,cdots,x_n) = prod_{i}x_i(1-x_i) prod_{i
Suppose all $$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.

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
$$D_i=frac 1{x_i}-frac 1{1-x_i}+sum_{j:jne i}frac 1{x_i-x_j}$$
is positive.
Now consider
$$sum_i x_i(1-x_i)D_i=sum_i(1-2x_i)+sum_{i,j:ine j}frac {x_i-x_i^2}{x_i-x_j}=sigma+Sigma,.$$
We have $$sigma=n-2sum_ix_i>0$$ by the assumption. Now we can use the antisymmetry of the denominator $$x_i-x_j$$ to write
$$2Sigma=sum_{i,j:ine j}frac {(x_i-x_i^2)-(x_j-x_j^2)}{x_i-x_j} \ = sum_{i,j:ine j}(1-x_i-x_j)=n(n-1)-2(n-1)sum_i x_i>0$$
and we are done.

Is it just an exercise from some book or you really needed it for something?

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