Positive definite form

Decide whether \(g(\mathbf u)>0\) holds for all nontrivial \(\mathbf u=(x_1, x_2, x_3)\in \mathbb R^3\).

Variant
\(g(\mathbf u)=x_1^2+2x_1x_2+2x_1x_3+2x_2^2+5x_3^2\)
Solution
We build the matrix of the form

\( \mathbf A= \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 5 \end{pmatrix} \)

and check whether it is positive definite.

E.g. \(|\mathbf A|=3>0\), \(\begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} =1 > 0 \) and also \(|1|=1>0\).
Answer
Yes, \(g(\mathbf u)>0\) for all \(\mathbf u\ne \mathbf 0\).
Variant
\(g(\mathbf u)=x_1^2+2x_1x_2+2x_1x_3+2x_2^2+5x_3^2+2ax_2x_3\), solve w.r.t. the parameter \(a\).

Does for all other choices of \(a\) hold that \(g(\mathbf u)\le 0\) for all \(\mathbf u\)?
Solution
\( \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & a \\ 1 & a & 5 \end{vmatrix} = -a^2+2a+3>0 \) for \(a\in (-1, 3)\).

The remaining two determinants are the same as in the previous case and are positive.
Answer
It holds that \(g(\mathbf u)>0\) for \(a\in (-1, 3)\).

For \(a\not\in (-1, 3)\) and \(\mathbf e^1=(1, 0, 0)^T\) is \(g(\mathbf e^1)=1\), hence the form \(g\) may still attain positive values – the answer is negative.