已知数列 $\left\{a_{n}\right\}$ 满足 $a_{n+1}+\frac{1}{a_{n+1}}=2 a_{n}+\frac{1}{a_{n}}\left(n \in N^{*}\right)$, 则 ( )
答案: C

解: $a_{n+1}+\frac{1}{a_{n+1}}=2 a_{n}+\frac{1}{a_{n}} \therefore a_{n+1}-a_{n}+\frac{1}{a_{n+1}}-\frac{1}{a_{n}}=a_{n}$,

即 $\left(a_{n+1}-a_{n}\right)\left(1-\frac{1}{a_{n+1} a_{n}}\right)=a_{n}$,

当 $0<a_{n}<1$ 时, $1-\frac{1}{a_{n+1} a_{n}}<0$, 故 $a_{n+1}<a_{n}, \mathrm{~A}$ 不符合题意;

当 $a_{n}>1_{\text {时, }} 1-\frac{1}{a_{n+1} a_{n}}>0$, 故 $a_{n+1}>a_{n}, \mathrm{~B}$ 不符合题意;

对于 $\mathrm{D}$ 选项, 当 $n=1$ 时, $a_{1}=2, a_{2}+\frac{1}{a_{2}}=2 a_{1}+\frac{1}{a_{1}}=\frac{9}{2}<\sqrt{23}$, $\mathrm{D}$ 不符合题意;

用数学归纳法证明 C,

易知 $a_{n}>0$ 恒成立,

当 $n=1$ 时, $a_{2}+\frac{1}{a_{2}}=2 a_{1}+\frac{1}{a_{1}}=3>\sqrt{6}$, 成立;

假设当 $n=k$ 时成立, $a_{k+1}+\frac{1}{a_{k+1}}>\sqrt{2 k+4}$, 即 $a^{2}{ }_{k+1}+\frac{1}{a^{2}{ }_{k+1}}>2 k+2$,

当 $n=k+1$ 时:

$\left(a_{k+2}+\frac{1}{a_{k+2}}\right)^{2}=\left(2 a_{k+1}+\frac{1}{a_{k+1}}\right)^{2}=4 a^{2}{ }_{k+1}+\frac{1}{a^{2}{ }_{k+1}}+4=a^{2}{ }_{k+1}+\frac{1}{a^{2}{ }_{k+1}}+3 a^{2}{ }_{k+1}+4>2 k+6$

即 $a_{k+2}+\frac{1}{a_{k+2}}>\sqrt{2 k+6}$ 成立,

故 $a_{n+1}+\frac{1}{a_{n+1}}>\sqrt{2 n+4}$ 恒成立, 即可得证.

故答案为: C

【分析】依次判断每个选项的正误, 得到答案.