Comon's conjecture on tensor rank versus symmetric rank

Let $V$ be a vector space and let $p\in\mathbb P S^dV\vartriangleleft\mathbb P(V^{\otimes d})$. Let $R_{v_d(\mathbb P V)}(p)$ be the symmetric rank of $p$, and let $R_{\operatorname{Seg}(\mathbb P V\times\cdots\times\mathbb P V)}(p)$ be its tensor rank. Comon's conjecture.

$$R_{v_d(\mathbb P V)}(p)=R_{\operatorname{Seg}(\mathbb P V\times\cdots\times\mathbb P V)}(p).$$

The supplied text states this as Comon's conjecture and gives no resolution.