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 $\forall x,y,z \in \mathbb{R}$

One of the methods when you don't know necessary and sufficient condition for the minimum of function of several variables - consider other as parameters. You know that for a function $$ a_1x^2+b_1(y,z)x+c(y,z) $$ the minimum is attained at $\frac{-b_1(y,z)}{2a_1}$ for $a_1>0$ and any fixed $y,z$. Then you should just substitute this into your equation and solve the minimum problem w.r.t. $y$ and then, on the third step, for $z$.

In your case: $a_1 = 3, b_1 = 8y+5z$, so you put $$ x = -\frac{1}{6}(8y+5z) $$ and obtain a function $$ \frac{1}{12}(20 y^2-56 y z-z^2) $$ which certainly can go below zero due to the negativity of the coefficient with $z^2$.

Finally, the strict inequality never holds, since any quadratic function is equal to zero in the origin.