Conditions for a certain parametric matrix to be positive definite The task:

Let's consider matrix $A=\begin{bmatrix}1&a&a\\a&1&a\\a&a&1 \end{bmatrix}$. Show that $A>0$ if and only if $-1<2a<2$.

Solution attempt: By definition $A$ is positive definite iff $\forall_{z\in\mathbb{R}^3}z^{\operatorname{T}}Az>0$.
For $z=\begin{bmatrix}z_1\\z_2\\z_3\end{bmatrix}$, $$\left(z^{\operatorname{T}}A\right)z=\left(z_1^2+az_1z_2+az_1z_3\right)+\left(az_1z_2+z_2^2+az_2z_3\right)+\left(az_1z_3+az_2z_3+z_3^2\right)\\=z_1^2+z_2^2+z_3^2+2a\left(z_1z_2+z_1z_3+z_2z_3\right)>0\\\iff\\z_1^2+z_2^2+z_3^2>-2a\left(z_1z_2+z_1z_3+z_2z_3\right)$$
This is trivially true for $z_1z_2+z_1z_3+z_2z_3\geq0$, so without loss of generality let's assume that $z_1z_2+z_1z_3+z_2z_3<0$, and thus $-(z_1z_2+z_1z_3+z_2z_3)>0$. Therefore we need to prove that:
$$\left({\Large\forall_{z_1,z_2,z_3}}-\frac{z_1^2+z_2^2+z_3^2}{z_1z_2+z_1z_3+z_2z_3}>2a\right)\iff -1<2a<2$$
And this seems false to me. This is because $-\frac{z_1^2+z_2^2+z_3^2}{z_1z_2+z_1z_3+z_2z_3}$ is always $\geq0$. For this reason we have a trivial counterexample, eg when $2a=-1000$ then  $2a\not\in(-1;2)$ but nonetheless $\forall_{z_1,z_2,z_3}-\frac{z_1^2+z_2^2+z_3^2}{z_1z_2+z_1z_3+z_2z_3}>2a$.
I can't find my error; could you kindly help me please?


$z_1^2+z_2^2+z_3^2+2a\left(z_1z_2+z_1z_3+z_2z_3\right)>0\iff z_1^2+z_2^2+z_3^2>-2a\left(z_1z_2+z_1z_3+z_2z_3\right)$
This is trivially true for $z_1z_2+z_1z_3+z_2z_3\geq0$

Not, it is not, since $a$ might be negative.
I suppose showing $A$ is definite positive could be done your way, but it's way too exhausting for me. I'd go with using the fact that $$A\text{ definite positive } \iff \text{ its eigenvalues are } > 0$$
Now choose your favorite way to find the eigenvalues of $A$. Note that the sum of the rows are $2a+1$ so this is one of them ; which by the way leads to $-1<2a$. Now your turn...
