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?

Hint :
Using Fuchs's inequality wich is an extension of Karamata's inequality one can show that for ($f(x)=x^n$),$a\geq b \geq c>0$ such that $abc=1$:
$$\left(\frac{a^{n}}{\left(a+c\right)\left(a+c\right)}-1\right)a\cdot\frac{1}{\left(a-b\right)\left(a-c\right)}\geq \left(\frac{b^{n}}{\left(b+c\right)\left(b+a\right)}-1\right)\cdot\frac{b}{\left(b-a\right)\left(b-c\right)}\geq \left(\frac{c^{n}}{\left(a+c\right)\left(b+c\right)}-1\right)\cdot\frac{c}{\left(c-a\right)\left(c-b\right)}$$
We have the inequalities :
$$\left(\frac{a^{n}}{\left(a+c\right)\left(a+c\right)}-1\right)a\cdot\frac{1}{\left(a-b\right)\left(a-c\right)}\ge 0$$
$$\left(\frac{a^{n}}{\left(a+c\right)\left(a+c\right)}-1\right)a\cdot\frac{1}{\left(a-b\right)\left(a-c\right)}+\left(\frac{b^{n}}{\left(b+c\right)\left(b+a\right)}-1\right)\cdot\frac{b}{\left(b-a\right)\left(b-c\right)}\ge 0$$
And :
$$\left(\frac{a^{n}}{\left(a+c\right)\left(a+c\right)}-1\right)a\cdot\frac{1}{\left(a-b\right)\left(a-c\right)}+ \left(\frac{b^{n}}{\left(b+c\right)\left(b+a\right)}-1\right)\cdot\frac{b}{\left(b-a\right)\left(b-c\right)}+ \left(\frac{c^{n}}{\left(a+c\right)\left(b+c\right)}-1\right)\cdot\frac{c}{\left(c-a\right)\left(c-b\right)} \ge 0$$
So using Fuchs's inequality we have shown that :
$$\frac{a^{n}}{(a-b)(a-c)}+\frac{b^{n}}{(b-a)(b-c)}+\frac{c^{n}}{(c-a)(c-b)}\le \frac{a^{2n}}{(a^{2}-b^{2})(a^{2}-c^{2})}+\frac{b^{2n}}{(b^{2}-a^{2})(b^{2}-c^{2})}+\frac{c^{2n}}{(c^{2}-a^{2})(c^{2}-b^{2})}$$
Now conclude is easy in fact .
Hope it helps .
