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For integers $n>1$, show that
$$\left(\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n}{n+1}\right)^{n}>\frac{n^{n}}{n+1} .$$
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Applying AM-GM, we get that
$$\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n}{n+1} \geq n \sqrt[n]{\frac{1}{2} \times \frac{2}{3} \times \cdots \times \frac{n}{n+1}}=n \sqrt[n]{\frac{1}{n+1}}$$
Raising both sides to the $n^{\text {th }}$ power, we obtain
$$\left(\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n}{n+1}\right)^{n} \geq \frac{n^{n}}{n+1} .$$
It is obvious that the equality case does not apply since $\frac{1}{2} \neq \frac{2}{3} \cdot \square$
by Diamond (80,728 points)

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