Alexander’s conjecture on an inequality of Cassels

For every integer n≥1n\ge 1, every real number ρ>1\rho>1, and all complex numbers z1,…,znz_1,\ldots,z_n satisfying ∣zj∣≤ρ|z_j|\le \rho for 1≤j≤n1\le j\le n, prove that ∏1≤j,k≤nj≠k∣1−zj‾zk∣≤(ρ2n−1ρ2−1)n\displaystyle\prod_{\substack{1\le j,k\le n\\ j\ne k}}\left|1-\overline{z_j}z_k\right|\le \left(\frac{\rho^{2n}-1}{\rho^2-1}\right)^n.

References

Additional references

Progress summary

Refreshed
Claimed solved

A new paper claims to prove the conjecture completely, but the result has not been independently verified here.

Alexander’s conjecture concerns extending a product inequality of Cassels to all ρ>1\rho>1. Myriam Ounaïes claims that the conjecture holds in full, with equality only for a regular nn-gon on the boundary circle.

Known results

  • Cassels proved the inequality under the restriction cos⁡(π/n)≤ρ2/(ρ4−ρ2+1)\cos(\pi/n)\leq\rho^2/(\rho^4-\rho^2+1).
  • Alexander observed that this restriction could be weakened to cos⁡(π/n)≤2ρ2/(ρ4+1)\cos(\pi/n)\leq2\rho^2/(\rho^4+1) and conjectured validity for every ρ>1\rho>1.
  • Dubickas proved a related elementary-symmetric-function conjecture in degrees 11 through 44, which the paper says would imply Alexander’s conjecture.

September 2026 proof claim

Ounaïes’s preprint, also published in Complex Analysis and Operator Theory, claims the unrestricted inequality

∏j≠k∣1−zj‾zk∣≤(ρ2n−1ρ2−1)n\prod_{j\ne k}|1-\overline{z_j}z_k|\leq\left(\frac{\rho^{2n}-1}{\rho^2-1}\right)^n

\nfor ∣zj∣≤ρ|z_j|\leq\rho, with equality exactly at the vertices of a regular nn-gon on ∣z∣=ρ|z|=\rho. The claim is unverified from the supplied evidence.

Current status (as of September 2026): Alexander’s conjecture is claimed solved by Ounaïes, but the proof and its exact publication scope remain independently unverified.

Sources

Solutions 0

No solutions have been posted yet.