<s>[INST] Show that $f(x,y,z)=6x^2+4y^2+2z^2+4xz-4yz \geq 0$ for all $x, y, z$ except for $x=y=z=0$ When I try to factor the quadratic form, I end up with 
$$6x^2+4y^2+2z^2+4xz-4yz = 2((x+z)^2+2x^2+(y-z)^2-z^2)$$ 
which does not ensure that $f(x,y,z) \geq 0$ for all $x, y, z \geq 0$ since the $z$ term is negative. How should these kinds of problems be tackled?
 [/INST] Your quadratic form is positive definite. I do not know what the eigenvalues of the matrix are. This method is the same as "completing the square." The diagonal entries of the diagonal matrix $D$ are all positive.
$$ P^T H P = D  $$
$$\left( 
\begin{array}{rrr} 
1 & 0 & 0 \\ 
0 & 1 & 0 \\ 
 -  \frac{ 1 }{ 3 }  &  \frac{ 1 }{ 2 }  & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
6 & 0 & 2 \\ 
0 & 4 &  - 2 \\ 
2 &  - 2 & 2 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 & 0 &  -  \frac{ 1 }{ 3 }  \\ 
0 & 1 &  \frac{ 1 }{ 2 }  \\ 
0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrr} 
6 & 0 & 0 \\ 
0 & 4 & 0 \\ 
0 & 0 &  \frac{ 1 }{ 3 }  \\ 
\end{array}
\right) 
$$
$$ Q^T D Q = H  $$
$$\left( 
\begin{array}{rrr} 
1 & 0 & 0 \\ 
0 & 1 & 0 \\ 
 \frac{ 1 }{ 3 }  &  -  \frac{ 1 }{ 2 }  & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
6 & 0 & 0 \\ 
0 & 4 & 0 \\ 
0 & 0 &  \frac{ 1 }{ 3 }  \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 & 0 &  \frac{ 1 }{ 3 }  \\ 
0 & 1 &  -  \frac{ 1 }{ 2 }  \\ 
0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrr} 
6 & 0 & 2 \\ 
0 & 4 &  - 2 \\ 
2 &  - 2 & 2 \\ 
\end{array}
\right) 
$$
</s>