<s>[INST] Find the interval of $a^2+b^2+c^2$ 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 $a^2+b^2+c^2?$
 [/INST] 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!
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