Let $n\ge3$ be an integer, and denote $\sigma_1,\ldots,\sigma_n$ the elementary symmetric polynomials in $n$ indeterminates: $$\sigma_1(X)=X_1+\cdots+X_n,\quad\ldots\quad,\sigma_n(X)=X_1\cdots X_n.$$ Because $\sigma_n$ is hyperbolic in the direction of $e=(1,\ldots,1)^T$, and $\sigma_k$ is the derivative of $\sigma_{k+1}$ in the direction $e$ (up to a constant positive factor), each of the $\sigma_k$ is hyperbolic in the direction $e$. Moreover, the following set is a convex cone: $$\Gamma_{k,n}=\{x\in{\mathbb R}^n\,;\,\sigma_1(x)\ge0,\ldots,\sigma_k(x)\ge0\}.$$ Notice that $\Gamma_k$ is symmetric under the permutation of coordinates.
Obviously $\Gamma_{n,n}$, which is the positive orthant, is self-dual, when ${\mathbb R}^n$ is equipped with the standard scalar product $x\cdot y$. Likewise $\Gamma_{2,n}$, a circular cone, is self-dual, but for a modified scalar product $\langle x,y\rangle=x\cdot y+\left(\frac n2-1\right)\sigma_1(x)\sigma_1(y)$. Mind that $\langle \cdot,\cdot\rangle$ is also invariant under permutation of coordinates.
My question is whether every $\Gamma_{k,n}$ (for $k=2,\ldots,n$) is self-dual when ${\mathbb R}^n$ is equipped with a scalar product $$\langle x,y\rangle=x\cdot y+\alpha\sigma_1(x)\sigma_1(y)$$ for a suitable parameter $\alpha$.
This parameter should equal $$-\inf\frac{x\cdot y}{\sigma_1(x)\sigma_1(y)}$$ where $x,y$ run over the non-zero elements of $\Gamma_{k,n}$. This number is presumably $\frac{n-k}{n(k-1)}$.
