Denote $E(a,b,c)=\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}; F(a,b,c)=\dfrac{a}{c}+\dfrac{c}{b}+\dfrac{b}{a}$. $\min{\left(\dfrac{1}{E(a,b,c)}-\dfrac{2}{F(a,b,c)}\right)}=-\max{\left(\dfrac{2}{F(a,b,c)}-\dfrac{1}{E(a,b,c)}\right)}$. Hold the properties: $E(a,b,c)=E(b,c,a)=E(c,a,b)=F(a,c,b)=F(c,b,a)=F(b,a,c)$; $E(ka,kb,kc)=E(a,b,c);F(ka,kb,kc)=F(a,b,c),\forall a,b,c,k>0$. For $a,b,c$ fixed numbers, but in different order, $\max{\left(\dfrac{2}{F(a,b,c)}-\dfrac{1}{E(a,b,c)}\right)}$ occurs for $F(a,b,c)\le E(a,b,c)$. WLOG, we can assume $\min{\{a,b,c\}}=a$. Using the previous properties: $E(a,b,c)=E(1,x,y); F(a,b,c)=F(1,x,y)$, where $x=\dfrac{b}{a},y=\dfrac{c}{a}$. $F(a,b,c)\le E(a,b,c)\Longleftrightarrow F(1,x,y)\le E(1,x,y)\Longleftrightarrow x\le y$. For fixed $y\ge1$ and $x$ variable, $1\le x\le y$, denote $m(x)=F(1,x,y),M(x)=E(1,x,y)$. $\dfrac{2}{m(x)}-\dfrac{1}{M(x)}=\dfrac{2-\dfrac{m(x)}{M(x)}}{m(x)}$. $m(x)=\dfrac{1}{y}+\dfrac{y}{x}+x; \dfrac{1}{y}+2\sqrt{y}\le m(x)$ (equality for $x=\sqrt{y}$) and $m(x)\le 1+y+\dfrac{1}{y}$ (equality for $x=1$ or $x=y$). $M(x)=\dfrac{1}{x}+\dfrac{x}{y}+y; y+\dfrac{2}{\sqrt y}\le M(x)$ (equality for $x=\sqrt{y}$) and $M(x)\le 1+y+\dfrac{1}{y}$ (equality for $x=1$ or $x=y$). After calculus, results: $\dfrac{m(x)}{M(x)}\ge \dfrac{m(\sqrt y)}{M(\sqrt y)}\Longleftrightarrow (y^3-1)(x-\sqrt y)^2\ge0$, true. Hence, $\max\left(\dfrac{2}{m(x)}-\dfrac{1}{M(x)}\right)=\dfrac{2}{m(\sqrt y)}-\dfrac{1}{M(\sqrt y)}=$ $=\dfrac{2t^5-2t^4+4t^2-t}{2t^6+5t^3+2}$, where $t=\sqrt y$. $\dfrac{2}{m(1)}-\dfrac{1}{M(1)}=\dfrac{1}{3}$. $\dfrac{2}{m(\sqrt y)}-\dfrac{1}{M(\sqrt y)}-\dfrac{1}{3}=-\dfrac{(t-1)^2(2t^4-2t^3+7t+2)}{3(2t^6+5t^3+2)}\le0$ for $t\ge 1$, with equality for $t=1$. Results: $\max{\dfrac{2}{m(\sqrt y)}-\dfrac{1}{M(\sqrt y)}}=\dfrac{1}{3}$, for $x=y=1\Longleftrightarrow a=b=c$. Final result: $\min{\left(\dfrac{1}{E(a,b,c)}-\dfrac{2}{F(a,b,c)}\right)}=-\dfrac{1}{3}$, for $a=b=c$.

$\frac{a}{c}=x,\frac{c}{b}=y,\frac{b}{a}=z$ and $x+y+z \geq 3$ $\frac{1}{xy+yz+zx}-\frac{2}{x+y+z}\geq \frac{3}{(x+y+z)^2}-\frac{2}{x+y+z} =\frac{(3-(x+y+z))^2}{3(x+y+z)^2}-\frac{1}{3} \geq -\frac{1}{3}$ with equality for $x=y=z=1 \to a=b=c $