System of equations with inequality

Question

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) $abc\gt-1$

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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.

Answer 1

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

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