Show that $2Q(x)$ can be written as the sum of three perfect squares

I will put the Hessian matrix of $x^2 + y^2 + z^2 - yz - zx -xy$ and display , with square matrices $PQ = QP = I,$ $P^T HP = D$ is diagonal, so $Q^T D Q = H$

I will type in the final outcome first. The $Q^T DQ$ version reads $$ \left( x - \frac{y}{2} - \frac{z}{2} \right)^2 + \frac{3}{4} (y-z)^2 = x^2 + y^2 + z^2 - yz - zx - xy $$

which does show that the quadratic form is rank two, not full rank. Furthermore, the form is zero when $x=y=z$ and only then. This does suggest that we might try to write the form using $(u-v)^2$ and $(v-w)^2.$ It will work out better if $(w-u)^2$ is included somehow $$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

$$ P^T H P = D $$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 \\ 1 & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & - 1 & - 1 \\ - 1 & 2 & - 1 \\ - 1 & - 1 & 2 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & \frac{ 3 }{ 2 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ $$

$$ Q^T D Q = H $$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 2 } & 1 & 0 \\ - \frac{ 1 }{ 2 } & - 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & \frac{ 3 }{ 2 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & - 1 & - 1 \\ - 1 & 2 & - 1 \\ - 1 & - 1 & 2 \\ \end{array} \right) $$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$