Math Genius: Change along some direction is positive

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$$f(x_1,cdots,x_n) = prod_{i}x_i(1-x_i) prod_{i<j}|x_i-x_j|$$
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.

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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