A question involving inequality

Question

$x=cy+bz\ ,\ y=az+cx$ 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 $abc>-1$.

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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.

$$ \begin{bmatrix} -1 & c & b \\ c & -1 & a\\ b & a & -1 \\ \end{bmatrix} $$

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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

Answer 1

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

Answer 2

Your result is correct. Also note that if three real numbers satisfy:$a^2+b^2+c^2+2abc=1$, then either $0 <|a|, |b|, |c| <1$ or $|a|, |b|, |c| >1$, the converse may or may not be true.