# If $abc=1$ ,and $n$ is a natural number,prove $\frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} \geqslant \frac{n(n-1)}{2}$

inequality

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:

$$P_2 = 1 \geq 1 = \frac12 \cdot 2 \cdot (1)\\ P_3 = a + b +c \geq 3 [abc]^{\frac{1}{3}} = 3 = \frac12 \cdot 3 \cdot(2)\\ P_4 = a^2 + b^2 + c^2 + ab + bc + ac \geq 6 [abc]^{\frac{2}{3}} = 6 = \frac12 \cdot 4 \cdot(3)$$