Recall. A second-order tensor is a bilinear form $$\label{1}\tag{1} B(x, y)=\sum_{i, j=1}^d b_{ij}x_iy_j, \qquad \text{where }x, y\in \mathbb R^d.$$ We will always assume, without loss of generality, that $B$ is symmetric; that is, $b_{ij}=b_{ji}$. We say that $B$ is positive semidefinite if $$\label{2}\tag{2} B(x,x)\ge 0, \qquad \text{for all }x\in\mathbb R^d.$$ There are many criteria to establish whether \eqref{2} holds.
I am interested in the analogous problem with an higher degree of multilinearity. Namely, consider the general third-order tensor $$ T(x, y, z)=\sum_{i,j,k=1}^d t_{ijk}x_iy_jz_k, $$ and the fourth-order tensor $$ F(x, y, z, w)=\sum_{i,j,k, h=1}^d f_{ijkh}x_iy_jz_kw_h. $$ We say that these are positive semidefinite if $$\label{3}\tag{3} T(x,x,x)\ge 0, \qquad \forall x\in \mathbb R^d\text{ with positive entries}, $$ and $$\label{4}\tag{4} F(x,x,x,x)\ge 0, \qquad \forall x\in\mathbb R^d.$$
Remark. In the third-order case, it is necessary to require that the entries of $x$ are all positive. Without that requirement, only the identically zero tensor $T=0$ would satisfy \eqref{3}.
My question is: which criteria are there to establish \eqref{3} and \eqref{4}?
