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Algebra Difficulty 6.0 AIME, harder Prove it

8.15. Prove that for any natural n⩾2n \geqslant 2 the following inequalities hold:

n(n+1n−1)<1+12+13+…+1n<n(1−1nn)+1 n(\sqrt[n]{n+1}-1)<1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}<n\left(1-\frac{1}{\sqrt[n]{n}}\right)+1

Solution

8.15. Apply the inequality between the arithmetic mean and the geometric mean for the numbers 2,32,43,…,n+1n2, \frac{3}{2}, \frac{4}{3}, \ldots, \frac{n+1}{n}. As a result, we get

1n(2+32+43+…+n+1n)>n+1n \frac{1}{n}\left(2+\frac{3}{2}+\frac{4}{3}+\ldots+\frac{n+1}{n}\right)>\sqrt[n]{n+1}

i.e.,

(1+1)+(1+12)+(1+13)+…+(1+1n)>nn+1n (1+1)+\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\ldots+\left(1+\frac{1}{n}\right)>n \sqrt[n]{n+1}

Apply the inequality between the arithmetic mean and the geometric mean for the numbers 1,12,23,…,n−1n1, \frac{1}{2}, \frac{2}{3}, \ldots, \frac{n-1}{n}. As a result, we get

1n(1+12+23+…+n−1n)>1nn \frac{1}{n}\left(1+\frac{1}{2}+\frac{2}{3}+\ldots+\frac{n-1}{n}\right)>\sqrt[n]{\frac{1}{n}}

i.e.,

1+(1−12)+(1−13)+…+(1−1n)>nnn 1+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\ldots+\left(1-\frac{1}{n}\right)>\frac{n}{\sqrt[n]{n}}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.