3元n次不等式

jhsinutopia

$a,b,c>0$, $a^2+b^2+c^2=1$, $n\in\mathbb{Z}, n>1$
Prove that $\frac{a}{1-a^n}+\frac{b}{1-b^n}+\frac{c}{1-c^n} \ge \frac{(n+1)^{1+\frac{1}{n}}}{n}$.

Aluminiumor

只需证
$$0<x<1,\frac{x}{1-x^n}\ge \frac{\left( n+1 \right) ^{1+\frac{1}{n}}}{n}x^2$$
即证
$$0<x<1,f\left( x \right) =x\left( 1-x^n \right) \le \frac{n}{\left( n+1 \right) ^{1+\frac{1}{n}}}$$
而
$$f'(x) =1-\left( n+1 \right) x^n$$
故
$$f\left( x \right) \le f\left( \frac{1}{\left( n+1 \right) ^{\frac{1}{n}}} \right) =\frac{1}{\left( n+1 \right) ^{\frac{1}{n}}}\cdot \frac{n}{n+1}=\frac{n}{\left( n+1 \right) ^{1+\frac{1}{n}}}$$
原不等式得证.