Determining if a quadratic polynomial is always positive

This is what Sylvester's criterion is for. Write your quadratic as $v^T A v$ where $v$ is a vector of variables $(x_1\ x_2\ \cdots\ x_n)$ and $A$ is a matrix of constants. For example, in your case, you are interested in $$\begin{pmatrix} x & y & z \end{pmatrix} \begin{pmatrix} 3 & 4 & 5/2 \\ 4 & 7 & 1 \\ 5/2 & 1 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}$$ Observe that the off diagonal entries are half the coefficients of the quadratic.

The standard terminology is that $A$ is "positive definite" if this quantity is positive for all nonzero $v$. Sylvester's criterion says that $A$ is positive definite if and only if the determinants of the top-left $k \times k$ submatrix are positive for $k=1$, $2$, ..., $n$. In our case, we need to test $$\det \begin{pmatrix} 3 \end{pmatrix} =3 \quad \det \begin{pmatrix}3 & 4 \\ 4 & 7\end{pmatrix} = 5 \quad \det \begin{pmatrix} 3 & 4 & 5/2 \\ 4 & 7 & 1 \\ 5/2 & 1 & 2 \end{pmatrix} = -67/4.$$ Since the last quantity is negative, Sylvester's criterion tells us that this quadratic is NOT positive definite.

Rewrite your expression as a bilinear form with a symmetric matrix in-between. This can always be done. For instance, in your case, your expression is $$3x^{2}+8xy+5xz+2yz+7y^{2}+2z^{2} = \begin{pmatrix} x & y & z \end{pmatrix} \begin{pmatrix} 3 & 4 & 5/2 \\ 4 & 7 & 1 \\ 5/2 & 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}$$ Now all you need to check is that the matrix is positive definite. A nice property of the positive definite matrix is that every diagonal sub-matrix must be positive definite. However, note that the matrix $$\begin{pmatrix} 3 & 5/2 \\ 5/2 & 2 \end{pmatrix}$$ is not positive definite. Hence, it is not possible that $$3x^{2}+8xy+5xz+2yz+7y^{2}+2z^{2}$$ is always positive $\forall x,y,z \in \mathbb{R}$