若 ${a_1}=1$，${a_{n+1}} \cdot {a_n}=n+1$，求证：$\dfrac{1}{{{a_1}}}+\dfrac{1}{{{a_2}}}+\cdots+\dfrac{1}{{{a_n}}} \geqslant 2\left( {\sqrt {n+1}-1} \right)$．
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【答案】
略
【解析】
根据题意$$\begin{aligned}
{a_{n+1}} \cdot {a_n}&=n+1, \\
{a_{n+2}} \cdot {a_{n+1}}&=n+2, \\
\end{aligned}$$于是$${a_{n+1}}\left( {{a_{n+2}}-{a_n}} \right)=1,$$也即$$\dfrac{1}{{{a_{n+1}}}}={a_{n+2}}-{a_n},$$于是\[\begin{split}
\sum\limits_{k=1}^n {\dfrac{1}{{{a_k}}}} &= \dfrac{1}{{{a_1}}}+\left( {{a_3}-{a_1}} \right)+\left( {{a_4}-{a_2}} \right)+\cdots+\left( {{a_{n+1}}-{a_{n-1}}} \right)\\
&= \dfrac{1}{{{a_1}}}+{a_n}+{a_{n+1}}-{a_1}-{a_2}\\
& \geqslant 1-1-2+2\sqrt {{a_n}{a_{n+1}}}\\
&= 2\left( {\sqrt {n+1}-1} \right),\end{split}\]于是原不等式成立．
{a_{n+1}} \cdot {a_n}&=n+1, \\
{a_{n+2}} \cdot {a_{n+1}}&=n+2, \\
\end{aligned}$$于是$${a_{n+1}}\left( {{a_{n+2}}-{a_n}} \right)=1,$$也即$$\dfrac{1}{{{a_{n+1}}}}={a_{n+2}}-{a_n},$$于是\[\begin{split}
\sum\limits_{k=1}^n {\dfrac{1}{{{a_k}}}} &= \dfrac{1}{{{a_1}}}+\left( {{a_3}-{a_1}} \right)+\left( {{a_4}-{a_2}} \right)+\cdots+\left( {{a_{n+1}}-{a_{n-1}}} \right)\\
&= \dfrac{1}{{{a_1}}}+{a_n}+{a_{n+1}}-{a_1}-{a_2}\\
& \geqslant 1-1-2+2\sqrt {{a_n}{a_{n+1}}}\\
&= 2\left( {\sqrt {n+1}-1} \right),\end{split}\]于是原不等式成立．
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