If $abc=1$ ,and $n$ is a natural number,prove $ \frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} \geqslant \frac{n(n-1)}{2} $ If $a, b, c$ are distinct positive real numbers such that $abc=1$,and $n$ is a natural number,prove
$$
\frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} \geqslant \frac{n(n-1)}{2}
$$
I know for $n=3$,the answer is here,and how to go further?

As can be found here and  here , we have that
$$
P_n = \frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} = 
\displaystyle\sum_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3}
$$
Now by AM-GM, it follows that
$$
P_n = 
\displaystyle\sum_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3}
\geq Q [\displaystyle\prod_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3} ]^{1/Q}
$$
where $Q$ is the number of terms in the sum $\sum_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3}$.
Now note that by symmetry,
$$
P_n 
\geq Q [\displaystyle\prod_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3} ]^{1/Q} = Q [a b c ]^{\frac{Q(n-2)}{3 Q}} = Q
$$
where the last one is true by the condition $abc =1 $. So it remains to find $Q$. This derives from the sum's conditions: we can select $n_1$ from $0$ to $n-2$, i.e. $n-1$ possibilites. Then we have that  $n_2$ can be taken from  $0$ to $n-2-n_1$, i.e. $n-1-n_1$ possibilites. So $Q = \sum_{n_1 =0}^{n-2} (n-1-n_1) = \frac12 n (n-1)$ which proves the claim. $\qquad \Box$
Some examples, using AM-GM directly:
$$P_2 = 1 \geq 1 = \frac12 \cdot 2 \cdot (1)\\
P_3 = a + b  +c \geq 3 [abc]^{\frac{1}{3}} = 3 = \frac12 \cdot 3 \cdot(2)\\
P_4 = a^2 + b^2 + c^2 + ab + bc + ac \geq 6 [abc]^{\frac{2}{3}} = 6 = \frac12 \cdot 4 \cdot(3)
$$
