$1 < a^n < b^n ⇒ 1 < a < b$ の証明

$a^n < b^n \Rightarrow \left( \frac{a}{a} \right)^n = 1 < \left( \frac{b}{a} \right)^n$

$1 < a^{1/n} < b^{1/n}$ ならば　$1 < a^{\frac{1}{n+1}} < b^{\frac{1}{n+1}}$ が言えればよい。

$1 < a^{1/n} < b^{1/n}\Rightarrow a^{1/n}a^{\frac{n-1}{n}} < b^{1/n}a^{\frac{n-1}{n}} < b^{1/n}a^{\frac{n-1}{n}}\frac{b^\frac{n-1}{n}}{a^\frac{n-1}{n}} = b^{1/n}a^{\frac{n-1}{n}}{\frac{b}{a}^\frac{n-1}{n}}$

$b^{\frac{n+1}{n}}\left( b^{\frac{1}{n+1}} - a^{\frac{1}{n+1}}\right) = b^{\frac{1}{n}}-bb^{\frac{1}{n}}a^{\frac{1}{n+1}} < b^{\frac{1}{n}}-a^{\frac{1}{n}}$ 　⇒　ダメ？

$b^{\frac{n+1}{n}}\left( b^{\frac{1}{n+1}} - a^{\frac{1}{n+1}}\right) = b^{\frac{1}{n}}-bb^{\frac{1}{n}}a^{\frac{1}{n+1}} > a^{\frac{n+1}{n}}\left( b^{\frac{1}{n+1}} - a^{\frac{1}{n+1}}\right) = aa^{\frac{1}{n}}b^{\frac{1}{n+1}}-a^{\frac{1}{n}}$　⇒　ダメ？

$a^{1/n} < b^{1/n}$　なので、 $a^{\frac{n}{n+1}}a^{1/n} < a^{\frac{n}{n+1}}
b^{1/n}$

b^{\frac{1}{n}}-a^{\frac{1}{n}}$