戴老师的一道数列不等式问题

(数列不等式)
已知数列$\{a_n\}$满足$3a_1=1,n^2a_{n+1}-a_n^2=n^2a_n\ (n\in \mathbb{N}^\ast)$,则下列选项正确的是 ( )

(A) $\{a_n\}$是递减数列

(B) $\{a_n\}$是递增数列,且存在$n\in \mathbb{N}^\ast$使得$a_n>1$

(C) $\displaystyle\frac{1}{a_{n+1}}>\frac{3}{2}$

(D) $\displaystyle a_{n+1}>\frac{n+1}{2n+3}$

证.由$a_1=\frac{1}{3}$得$a_2=\frac{4}{9}$.

由$n^2a_{n+1}-a_n^2=n^2a_n$可得
$$a_{n+1}=\frac{a_n\left( a_n+n^2 \right)}{n^2}=a_n\left( \frac{a_n}{n^2}+1 \right) >a_n,$$
则$a_n>0$且$\{a_n\}$是递增数列.

取倒数后得
$$\frac{1}{a_{n+1}}=\frac{n^2}{a_n\left( a_n+n^2 \right)}=\frac{1}{a_n}-\frac{1}{a_n+n^2}>\frac{1}{a_n}-\frac{1}{n^2},$$
故
$$\begin{aligned} \frac{1}{a_n} &=\left( \frac{1}{a_n}-\frac{1}{a_{n-1}} \right) +\cdots +\left( \frac{1}{a_3}-\frac{1}{a_2} \right) +\frac{1}{a_2} \\ &>\frac{1}{a_2}-\left( \frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{\left( n-1 \right) ^2} \right) \\ &>\frac{9}{4}-\left( \frac{1}{2^2}+\frac{1}{2\times 3}+\cdots +\frac{1}{\left( n-2 \right) \left( n-1 \right)} \right) \\ &=\frac{9}{4}-\frac{1}{2^2}-\frac{1}{2}+\frac{1}{n-1} =\frac{3}{2}+\frac{1}{n-1}>\frac{3}{2}. \end{aligned}$$
故C选项正确.

由$\displaystyle\frac{1}{a_n}>\frac{3}{2}$可得$\displaystyle a_n<\frac{2}{3}<1$,故B选项错误.

最后利用数学归纳法证明$\displaystyle a_n>\frac{n}{2n+1}\ (n\geqslant 2)$.

当$n=2$时, $\displaystyle a_2=\frac{4}{9}>\frac{2}{5}$成立.

假设$n=k$时结论成立,则$\displaystyle a_k>\frac{k}{2k+1}$,则
$$\begin{aligned} a_{k+1} &=a_k\left( \frac{a_k}{k^2}+1 \right) >\frac{k}{2k+1}\left( \frac{1}{k^2}\cdot \frac{k}{2k+1}+1 \right) \\ &=\frac{k}{2k+1}\left( \frac{1}{k\left( 2k+1 \right)}+1 \right) =\frac{k}{2k+1}+\frac{1}{\left( 2k+1 \right) ^2}=\frac{2k^2+k+1}{\left( 2k+1 \right) ^2}. \end{aligned}$$
接下来证明 $$\frac{2k^2+k+1}{\left( 2k+1 \right) ^2}>\frac{k+1}{2k+3},$$
等价于证明
$$\left( 2k^2+k+1 \right) \left( 2k+3 \right) >\left( k+1 \right) \left( 2k+1 \right) ^2,$$
即$4k^3+8k^2+5k+3>4k^3+8k^2+5k+1$,显然成立.

故$\displaystyle a_n>\frac{n}{2n+1}\ (n\geqslant 2)$,即$\displaystyle a_{n+1}>\frac{n+1}{2n+3}$.故D选项正确.

答案选CD.

证法二.或者由$a_n<1$可得
$$\frac{1}{a_{n+1}}=\frac{1}{a_n}-\frac{1}{a_n+n^2}<\frac{1}{a_n}-\frac{1}{n^2+1}<\frac{1}{a_n}-\frac{1}{n\left( n+1 \right)},$$
故
$$\begin{aligned} \frac{1}{a_n}&=\left( \frac{1}{a_n}-\frac{1}{a_{n-1}} \right) +\cdots +\left( \frac{1}{a_2}-\frac{1}{a_1} \right) +\frac{1}{a_1}\\ &<\frac{1}{a_1}-\left( \frac{1}{1\times 2}+\frac{1}{2\times 3}+\cdots +\frac{1}{n\left( n+1 \right)} \right)\\ &<3-\left( 1-\frac{1}{n+1} \right)\\ &=2+\frac{1}{n+1}<2+\frac{1}{n}=\frac{2n+1}{n}. \end{aligned}$$
于是$\displaystyle a_n>\frac{n}{2n+1}\ (n\geqslant 2)$.