<s>[INST] System of equations with inequality 
Question Statement:-
If $$\begin{eqnarray}
x=cy+bz\\
y=az+cx\\
z=bx+ay
\end{eqnarray}$$ where $x,y,z$ are not all zero, prove that $a^2+b^2+c^2+2abc=1$
Further if at least one of $a,b,c$ is a proper fraction, prove that:- 
(i) $a^2+b^2+c^2\lt 3$
(ii) $abc\gt-1$


Attempt at a solution:-
The first part (i.e the part where we have to prove $a^2+b^2+c^2+2abc=1$) of the question gets completed pretty easily.
As the given system of equations is homogeneous and as $x,y,z$ are not all zero so the solutions are non-trivial. And for a homogeneous system of equations to have non-trivial solutions we have :-
$$\begin{vmatrix}
1 & -c & -b\\
-c & 1 & -a\\
-b & -a & 1\\
\end{vmatrix}=0$$
And on expanding the determinant we get
$(1-a^2)+c(-c-ab)-b(ac+b)=0\implies a^2+b^2+c^2+2abc=1$
I am not able to come up with any way to solve the remaining two portions of the questions. 
 [/INST] Assuming $a$ is a proper fraction, then
$$a^2<1$$
Considering quadratic in $c$,
$$c^2+2abc+a^2+b^2-1=0$$
Admitting real value of $c$,
\begin{align*}
  \Delta & \ge 0 \\
  4(a^2-1)(b^2-1) &\ge 0 \\
  b^2-1 & \le 0
\end{align*}
Hence,
$$b^2 \le 1$$
Similarly,
$$c^2 \le 1$$
Hence,

$$a^2+b^2+c^2<3$$

Also

$$abc=\frac{1-a^2-b^2-c^2}{2}>-1$$

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