对任意正整数 $n$，设 $a_n$ 是方程 $x^2+\dfrac xn=1$ 的正根．
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【出处】
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【标注】
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求证：$a_{n+1}>a_n$；标注答案略解析由题意知 $a_n>0$，所以$$1=a_n^2+\dfrac {a_n}{n}>a_n^2+\dfrac {a_n}{n+1},$$从而有$$a_{n+1}^2+\dfrac {a_{n+1}}{n+1}=1>a_n^2+\dfrac {a_n}{n+1},$$整理得$$(a_{n+1}-a_n)\left(a_{n+1}+a_n+\dfrac {1}{n+1}\right )>0,$$从而有 $a_{n+1}>a_n$．
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求证：$\dfrac{1}{2a_2}+\dfrac{1}{3a_3}+\cdots +\dfrac{1}{na_n}<1+\dfrac 12+\dfrac 13+\cdots +\dfrac 1n$．标注答案略解析由已知可得 $na_n=\dfrac{\sqrt{4n^2+1}-1}2$，$n\in\mathbb N^*$．于是\[\begin{split} \dfrac{1}{na_n}-\dfrac 1n&=\dfrac{2}{\sqrt{4n^2+1}-1}-\dfrac 1n=\dfrac{\sqrt{4n^2+1}+1}{2n^2}-\dfrac 1n\\ &=\dfrac{\sqrt{4n^2+1}-2n+1}{2n^2}=\dfrac{\frac{1}{\sqrt{4n^2+1}+2n}+1}{2n^2}\\ &<\dfrac{1}{8n^3}+\dfrac{1}{2n^2} <\dfrac{1}{8(n-1)n(n+1)}+\dfrac{1}{2\left(n-\frac 12\right)\left(n+\frac 12\right)}\\ &=\dfrac 1{16}\left[\dfrac{1}{(n-1)n}-\dfrac{1}{n(n+1)}\right]+\dfrac 12\left(\dfrac{1}{n-\frac 12}-\dfrac{1}{n+\frac 12}\right) ,\end{split}\]因此$$\left(\dfrac{1}{2a_2}+\dfrac{1}{3a_3}+\cdots +\dfrac{1}{na_n}\right)-\left(\dfrac 12+\dfrac 13+\cdots +\dfrac 1n\right)<\dfrac{1}{16}\cdot\dfrac 12+\dfrac 12\cdot \dfrac 23=\dfrac{35}{96}<1,$$于是原不等式得证．
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