I tried homogenising and then isolated fudging. But while doing so I'm stuck at this step, if someone can help $$\left(\frac{ab+bc+ca}{3}\right)^{n+1}\geq \frac{3(bc)^{n+1}(a^{3n}+b^{3n}+c^{3n})}{a^{3n}+a^{n-1}bc(b^{2n-1}+c^{2n-1})}$$Note- this result is independent of the given condition, as we have already homogenised it

By the AM-GM on the condition we get $abc \ge 27,$ so $a^{n}+b^{n}+c^{n} \ge 3^{n+1},$ and by the Cauchy-Schwarz we have $(a^{2n+1}+bc^{2n}+cb^{2n})=(a^{2n+1}+bc^{2n}+cb^{2n}) * (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \ge (a^{n}+b^{n}+c^{n})^{2},$ and similarly on other two fractions we get that the answer is $\frac{1}{3^{n+1}}$ at $3,3,3.$