8．已知三个互异的正实数$a$，$b$，$c$满足$c = 2\ln\frac{c}{a} + a$，$b = \log_{\frac{\sqrt{5}}{5}}\frac{1}{2^{a} + 1}$，则关于$a$，$b$，$c$下列判断正确的是（ ）
A．$a < b < c$ B．$a > b > c$ C．$\left| a - c \right| < \left| b - 2 \right|$ D．$\left| a - c \right| > \left| b - 2 \right|$
解 $\because c = 2\ln\frac{c}{a} + a$
$\therefore c - 2\ln c = a - 2\ln a$，$\sqrt{5}^{b} = 2^{a} + 1$
设$f(x) = x - 2\ln x$，则$f'(x) = 1 - \frac{2}{x} = \frac{x - 2}{x}$
$\therefore x > 2$时$f(x)$单增，$x < 2$时$f(x)$单减
$\because f(c) = f(a)$
$\therefore a$与$c$必然一个大于2，一个小于2
若$0 < a < 2$，$c > 2$，则$\sqrt{5}^{b} = 2^{a} + 1 < 5$，$b < 2$
设$g(x) = \left( \frac{2}{\sqrt{5}} \right)^{x} + \left( \frac{1}{\sqrt{5}} \right)^{x}$，显然$g(x)$单减且$g(2) = 1$
$\therefore g(a) = \left( \frac{2}{\sqrt{5}} \right)^{a} + \left( \frac{1}{\sqrt{5}} \right)^{a} = \left( 2^{a} + 1 \right)\left( \frac{1}{\sqrt{5}} \right)^{a} = \sqrt{5}^{b - a} > g(2) = 1$
$\therefore b > a$
$\therefore 0 < a < b < 2 < c$
$\therefore \left| a - c \right| > \left| b - 2 \right|$
同理，若$0 < c < 2$，$a > 2$时可得$0 < c < b < 2 < a$
$\therefore \left| a - c \right| > \left| b - 2 \right|$ 故选 D