If $x= cy + bz$, $y= az + cx$, $z= bx + ay$ where $x, y, z$ are not all zeros and at least one of $a, b, c$ is a proper fraction. Prove following results:
$$\begin{align*}&(i)\quad\:\: a^{2} + b^{2} + c^{2} + 2abc= 1\\&(ii)\quad\:a^{2} + b^{2} + c^{2}\lt 3\\&(iii)\quad a\cdot b\cdot c\gt -1\end{align*}$$ I succeeded in proving the first result by setting $$\begin{vmatrix} 1 & -c & -b \\ -c & 1 & -a \\ -b & - a & 1\end{vmatrix}= 0$$ But I don't have a clue on how to show the next two results any help would be appreciated.
If $xyz = 0$, it is easy. Omitted.
In the following, assume that $xyz \ne 0$.
From $x = cy + bz, y = az + cx, z = bx + ay$, it is easy to obtain $$a = \frac{y^2 + z^2 - x^2}{2yz}, \quad b = \frac{z^2 + x^2 - y^2}{2zx}, \quad c = \frac{x^2 + y^2 - z^2}{2xy}.$$
It is easy to verify that $$(1 - a^2)\cdot 4y^2z^2 = (1 - b^2)\cdot 4z^2x^2 = (1 - c^2)\cdot 4x^2y^2.$$
Since at least one of $|a|, |b|, |c|$ is less than one, we have $1 - a^2 > 0, 1 - b^2 > 0, 1 - c^2 > 0$. Thus, $a^2 + b^2 + c^2 < 3$, and $abc > -1$.
We are done.