Consider the Vandermonde product $\prod\_{1\le j < k \le n} |z\_j - z\_k|$. It is well-known that under the constraint $|z\_j| \le 1$ for all $j$, the product is maximized at a picket fence configuration of the form $z\_j = e^{2 \pi i j / n}$, with a maximum value of $n^{n/2}$; see [here](http://mathworld.wolfram.com/HadamardsMaximumDeterminantProblem.html).


Now instead of the simple constraint above, I impose $|z\_j| \le r\_j$, with the convention that $0 < r\_1 \le r\_2 \le \ldots \le r\_n$. The maximizing configuration is obviously going to be more complicated, but I conjecture the following upper bound on the maximum *value*:
$$ \prod\_{1 \le j < k \le n}|z\_j - z\_k| \le n^{n/2}\prod\_{j=1}^n r\_j^{j-1}.$$


Is this true? Any partial result (including weaker bound) is welcome. 


Update: the case $n=2$ is trivially verified since the left hand side is $\le r\_1 + r\_2 \le 2 r\_2$. For $n=3$, consider projecting the **optimizing** $z\_j$'s radially to the outermost circle of radius $r\_3$, and call them $\zeta\_j$'s, so that $\zeta\_3 = z\_3$. One must have $|\zeta\_j - \zeta\_k| > r\_3$. This ensures that the radial projection can only increase the product. But for higher $n$, this last condition does not necessarily hold, so one has to account for the tradeoff between the losses and gains from outward radial projection.



An extension of Hadamard maximum determinant problem
Due to homogeneity, we assume that $r\_n=1$.


Set $B=\{z\in \colon |z|\leq 1\}$ and $T=\{z\colon |z|=1\}$. Consider the function 
$$
 f(z\_1,z\_2,\dots,z\_n)=\frac
 {\displaystyle \prod\_{1\leq i<j\leq n}(z\_i-z\_j)}
 {\displaystyle \prod\_{i=1}^n z\_i^{i-1}}
$$
on the set $S=\{(z\_1,\dots,z\_n)\in B^n\colon 0<|z\_1|\leq\dots\leq |z\_n|\}$ (yes, this set is not closed). We claim that the maximum of $|f|$ is attained at a point with $|z\_i|=1$ for all $i$ (that is, at a point in $T^n$); the required result will follow, since this reduces the problem to the case of equal radii.


For any point $\mathbf z=(z\_i)\in S$, set $\mu(\mathbf z)=\max\{m\leq n\colon |z\_1|=|z\_2|=\dots=|z\_m|\}$. To prove our claim, we will show that any point $\mathbf z=(z\_i)\in S\setminus T^n$ can be modified into a point $\mathbf w=(w\_i)\in S$ such that $|f(\mathbf w)|\geq |f(\mathbf z)|$ and $\mu(\mathbf w)>\mu(\mathbf z)$. After a chain of such modifications, we will arrive at a point with $\mu(\mathbf w)=n$, which is what we need (by homogeneity again).


So, assume that $\mu(\mathbf z)=m<n$. Set $t\_i=z\_i/z\_1$ for all $i\leq m$ (then $|t\_i|=1$), and consider the function $g(z)=f(zt\_1,zt\_2,\dots,zt\_m,z\_{m+1},\dots,z\_n)$ (so $g(z\_1)=f(\mathbf z)$). Notice that the whole power of $z$ in the denominator cancels out, so this function is a polynomial in $z$. Therefore, the maximum of absolute value of $g$ on $\{z\colon |z|\leq |z\_{m+1}|\}$ is attained at some point $z\_0$ with $|z\_0|=|z\_{m+1}|$, and we may set $\mathbf w=(z\_0t\_1,\dots,z\_0t\_m,z\_{m+1},\dots,z\_n)$. This finishes the proof.


