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cⁿ>aⁿ+bⁿ。
证明如下:
因a、b、c都是正数,且a²+b²=c²,即(a/c)²+(b/c)²=1且(a/c)<1、(b/c)<1。则:
(aⁿ)/(cⁿ)=(a/c)ⁿ=(a/c)^(n-2)(a/c)²<(a/c)²,同理,有:(bⁿ)/(cⁿ)<(b/c)²,所以,
(aⁿ)/(cⁿ)+(bⁿ)/(cⁿ)<(a/c)²+(b/c)²=1,即: (aⁿ)/(cⁿ)+(bⁿ)/(cⁿ)<1,就是cⁿ>aⁿ+bⁿ。
证明如下:
因a、b、c都是正数,且a²+b²=c²,即(a/c)²+(b/c)²=1且(a/c)<1、(b/c)<1。则:
(aⁿ)/(cⁿ)=(a/c)ⁿ=(a/c)^(n-2)(a/c)²<(a/c)²,同理,有:(bⁿ)/(cⁿ)<(b/c)²,所以,
(aⁿ)/(cⁿ)+(bⁿ)/(cⁿ)<(a/c)²+(b/c)²=1,即: (aⁿ)/(cⁿ)+(bⁿ)/(cⁿ)<1,就是cⁿ>aⁿ+bⁿ。
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