An extension of Hadamard maximum determinant problem 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.
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.
 A: 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.
