This is a wonderful problem.

Attachments:

Write $$\frac {a^{-n}+b}{1-a} = \frac {1 +a^{n}b}{a^n(b+c)}= \frac {a^{n+1} +1 +a^{n}b -a^{n+1}}{a^n(b+c)}= \frac {a+b}{b+c} + \frac {(b+c)(1+a+a^2+ \dots +a^n)}{a^n(b+c)}$$ So $$E = \sum_{\text {cyc}} \frac {a+b}{b+c} + 3 + \sum_{k=1}^{n} \left(\sum_{\text {cyc}} a^{-k} \right)$$ Now we use the estimates :== $$\sum_{\text {cyc}} \frac {a+b}{b+c} \overset {\text {AM-GM}}{\geq}3$$ And $$\left( \frac {a^{-k}+b^{-k} + c^{-k} }{3} \right) ^ {\frac {-1}{k}} \overset {\text {Power-Mean}}{\leq} \frac {a+b+c}{3} \implies a^{-k}+b^{-k}+c^{-k} \geq 3^{k+1}$$ So $$E \geq 3+3+3^2+ \dots + 3^{n+1} \implies \boxed{E_{\text {min}} = \frac {3^{n+2}+3}{2}}$$ Done . $\blacksquare$ Equality holds at $a=b=c= \frac 13$