If $g_j$ is SOS, i.e. $g_j=\sum_k h_k^2$, then $g_j(x)\leq 0$ iff $g_j(x)=0$, i.e. $h_1(x)=h_2(x)=\dots =0$. So this is a general case, although with equality constraints only.

A more interesting question would be to allow inhomogenous constraints $g_j(x)\leq b_j$, with $b_j\in\mathbb{R}$. IIRC, this would be solvable with just one step of the Lassere's hierarchy, as the whole point of it is approximating arbitrary nonnegative polynomials by SOS. (I believe the latest Lassere's book is discussing this in Chapter 11).

If $g_j$ is SOS, i.e. $g_j=\sum_k h_k^2$, then $g_j(x)\leq 0$ iff $g_j(x)=0$, i.e. $h_1(x)=h_2(x)=\dots =0$. So this is a general case, although with equality constraints only.

A more interesting question would be to allow inhomogenous constraints $g_j(x)\leq b_j$, with $b_j\in\mathbb{R}$. IIRC, this would be solvable with just one step of the Lassere's hierarchy, as the whole point of it is approximating arbitrary nonnegative polynomials by SOS. (I believe the latest Lasserre's book is discussing this in Chapter 11).

If $g_j$ is SOS, i.e. $g_j=\sum_k h_k^2$, then $g_j(x)\leq 0$ iff $g_j(x)=0$, i.e. $h_1(x)=h_2(x)=\dots =0$. So this is a general case, although with equality constraints only.

A more interesting question would be to allow inhomogenous constraints $g_j(x)\leq b_j$, with $b_j\in\mathbb{R}$. IIRC, this would be solvable with just one step of the Lassere's hierarchy, as the whole point of it is approximating arbitrary nonnegative polynomials by SOS.

If $g_j$ is SOS, i.e. $g_j=\sum_k h_k^2$, then $g_j(x)\leq 0$ iff $g_j(x)=0$, i.e. $h_1(x)=h_2(x)=\dots =0$. So this is a general case, although with equality constraints only.

A more interesting question would be to allow inhomogenous constraints $g_j(x)\leq b_j$, with $b_j\in\mathbb{R}$. IIRC, this would be solvable with just one step of the Lassere's hierarchy, as the whole point of it is approximating arbitrary nonnegative polynomials by SOS. (I believe the latest Lassere's book is discussing this in Chapter 11).