It is equivalent to $(\log_a b + \log_a c)^n + (\log_b a + \log_b c)^n + (\log_c a + \log_c b)^n \ge 3 \cdot 2^n$ by well-known log properties. The AM-GM inequality gives $LHS \ge 3 \sqrt[3]{(\log_a b + \log_a c)^n (\log_b a + \log_b c)^n (\log_c a + \log_c b)^n }$. It remains to show $(\log_a b + \log_a c)(\log_b a + \log_b c)(\log_c a + \log_c b) \ge 8$. With $x = \log_a b, y = \log_c a, xy = \log_c b$ we want to prove $y+xy+1+x+\frac{1}{x}+1+\frac{1}{xy}+\frac{1}{y} \ge 8$, trivial by AM-GM.

shmm wrote: Prove that : $\frac{1}{(\log_{bc} a)^n}+\frac{1}{(\log_{ac} b)^n}+\frac{1}{(\log_{bc} a)^n}\geq 3\cdot2^{n}$ where a,b,c>1 and n is natural number. Let $x=\log_{a} b,y=\log_{c} a,z=\log_{b} c$ , the problem is equivalent to Let $x,y,z$ be positive real numbers such that $xyz=1$. For the natural number $n$,prove that\[(x+\frac{1}{y})^n+(y+\frac{1}{z})^n+(z+\frac{1}{x})^n \geq3\cdot 2^n.\] See also here Croatian NMC 2005

$\frac{1}{(\log_{bc} a)^n}+\frac{1}{(\log_{ac} b)^n}+\frac{1}{(\log_{bc} a)^n}$ $=(\log_a b + \log_a c)^n + (\log_b a + \log_b c)^n + (\log_c a + \log_c b)^n $ $\ge 2^n\left((\log_a b \cdot \log_a c)^\frac{n}{2} + (\log_b a \cdot \log_b c)^\frac{n}{2} + (\log_c a \cdot \log_c b)^\frac{n}{2} \right)$ $\ge 3 \cdot 2^n \sqrt[3]{(\log_a b \cdot \log_a c)^\frac{n}{2} (\log_b a \cdot \log_b c)^\frac{n}{2} (\log_c a \cdot \log_c b)^\frac{n}{2} }=3 \cdot 2^n.$