Find the interval of $a^2+b^2+c^2$

Question

Given that the system of equations $x=cy+bz,y=az+cx,z=bx+ay$ has non-zero solutions and at least one of a,b,c is a proper fraction, then find the interval in which value of $a^2+b^2+c^2$ lies($a,b,c\in R$).

MY APPROACH:

I applied the cramer's rule and got the equation as $a^2+b^2+c^2=1-2abc$. Now how do i find the range of $a^2+b^2+c^2?$

Answer 1

Since one of the variables is a proper fraction, let us set, without loss of generality, $c = n/m$ where $0 < n < m$ and $n,m\in \mathbb{N}$. Then we have

$$a^2 + \dfrac{2n}{m}ab + \dfrac{n^2}{m^2} + b^2 - 1= 0.$$

Solving for $a$ we get

$$a = \dfrac{-\dfrac{2n}{m}b \pm \sqrt{\dfrac{4n^2}{m^2}b^2 - \dfrac{4n^2}{m^2} - 4b^2 + 4}}{2} = -\dfrac{n}{m}b \pm \sqrt{(1-b^2)\left(1 - \dfrac{n^2}{m^2}\right)}.$$

Then $-1 \leq b \leq 1$. Similarly, $-1\leq a \leq 1$ which would impose a strict upper bound of $a^2 + b^2 + c^2$ of $3$.

Indeed taking $m = n+1$ and making $n$ arbitrarilly large we can consider the solution

$$a = 1,\qquad b = -\dfrac{n}{n+1}, \qquad c = \dfrac{n}{n+1}$$

and

$$a^2 + b^2 + c^2 = 1 + 2\dfrac{n^2}{(n+1)^2},$$

which can be as close to $3$ as one desires.

Answer 2

Let $a=b=c=\frac{1}{2}$.

Hence, $a^2+b^2+c^2=\frac{3}{4}$.

We'll prove that $\frac{3}{4}$ is a minimal value.

Indeed, Let $a^2+b^2+c^2=\frac{3}{4}x^2$, where $x\geq0$.

Thus, by AM-GM $$1=a^2+b^2+c^2+2abc\leq\frac{3}{4}x^2+2|abc|\leq$$ $$\leq\frac{3}{4}x^2+2\left(\sqrt{\frac{a^2+b^2+c^2}{3}}\right)^3=\frac{3}{4}x^2+\frac{1}{4}x^3,$$ which gives $x\geq1$ and we are done!