If $a > b > 0,$ with the aid of Lagranges mean value theorem, prove that $n b^{n - 1} (a - b) 1 .$ $n b^{n - 1} (a - b) > a^{n} - b^{n} > n a^{n - 1} (a - b), if 0 < n < 1 .$

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a > b > 0,

with the aid of Lagranges mean value theorem, prove that

n b^{n - 1} (a - b) 1 .

n b^{n - 1} (a - b) > a^{n} - b^{n} > n a^{n - 1} (a - b), if 0 < n < 1 .

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Question Text

a > b > 0,

with the aid of Lagranges mean value theorem, prove that

n b^{n - 1} (a - b) 1 .

n b^{n - 1} (a - b) > a^{n} - b^{n} > n a^{n - 1} (a - b), if 0 < n < 1 .

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