My conjecture. Given three positive numbers $a, b, c$. There exists a positive non-cyclic polynomial $f\left ( a, b, c \right )$ so that $$f^{2}\left ( a, b, c \right )\geq f\left ( b, c, a \right )f\left ( c, a, b \right )$$
Is mine right ? I found that there exists a non-cyclic polynomial $f\left ( a, b, c \right )= a- b\neq 0$, which is $$\left ( a- b \right )^{2}\geq \left ( b- c \right )\left ( c- a \right )$$ because $$\left ( a- b \right )^{2}- \left ( b- c \right )\left ( c- a \right )= \left ( a+ b- 2c \right )^{2}+ 3\left ( b- c \right )\left ( c- a \right )\geq 0$$
Let there be such $f$.
Thus, we have $$f(a,b,c)^2\geq f(b,c,a)f(c,a,b)$$ and $$f(b,c,a)^2\geq f(c,a,b)f(a,b,c),$$ which gives $$f(a,b,c)^2f(b,c,a)^2\geq f(a,b,c)f(b,c,a)f(c,a,b)^2$$ or $$f(a,b,c)f(b,c,a)\left(f(a,b,c)f(b,c,a)-f(c,a,b)^2\right)\geq0$$ and since $$f(a,b,c)f(b,c,a)>0,$$ we obtain $$f(a,b,c)f(b,c,a)-f(c,a,b)^2\geq0,$$ which gives $$f(a,b,c)f(b,c,a)=f(c,a,b)^2,$$ Similarly $$f(a,b,c)f(c,a,b)=f(b,c,a)^2$$ and $$f(c,a,b)f(b,c,a)=f(a,b,c)^2.$$
The first and the second from three last equalities give: $$(f(a,b,c)+f(b,c,a)+f(c,a,b)(f(b,c,a)-f(c,a,b))=0$$ or $$f(b,c,a)=f(c,a,b)$$ and similarly we obtain: $$f(a,b,c)=f(b,c,a)=f(c,a,b),$$which gives that $f$ is a cyclic, which is a contradiction.
We see that this reasoning works for any positive function $f$ of three variables.