A question involving inequality $x=cy+bz\ ,\ y=az+cx$ and $z=bx+ay$ has a non zero solution and atleast one of a,b,c is a proper fraction(a rational number whose absolute value is less than 1) , prove that $a^3+b^3+c^3<3$ and $abc>-1$.

For non zero solution implies , the determinant of following marix be zero because if AX=B (A$^{-1}$ should not exist for infinite solution).And there is no possibility of no solution because we already know (0,0,0) is a solution.
$$ 
\begin{bmatrix}
-1 & c & b \\
c & -1 & a\\     
b & a & -1 \\
\end{bmatrix}
 $$

Giving us $a^2+b^2+c^2+2abc=1$. After this I am totally stuck. Somehow we have to use proper fraction condition .Thanks for the help

Let $|c|<1.$
Thus, $$1=a^2+b^2+c^2+2abc=(a+bc)^2+b^2+c^2-b^2c^2\geq b^2+c^2-b^2c^2,$$
which gives
$$(1-b^2)(1-c^2)\geq0$$ and $$b^2\leq1.$$
Similarly, $$a^2\leq1.$$
Thus, $$-1<abc<1$$ and $$a^3+b^3+c^3<3.$$
