# 构造函数的另一种思路：把两个未知中的其中一个看作函数自变量

## 二、解析

### 证明方法一：拉格朗日中值定理

$$f ( x ) = \ln x$$

$$\frac{\ln b−\ln a}{b−a} = \left(\ln x\right)^{\prime}\left|\right._{x = \xi } = \frac{1}{\xi }$$

$$\xi > 0, \ a > 0$$

$$a < \textcolor{orangered}{ \xi < b }$$

$$\frac{1}{a} > \textcolor{orangered}{ \frac{1}{\xi} > \frac{1}{b} }$$

$$\frac{1}{\xi} > \frac { 1 } { b } > \frac { 2 a } { a ^ { 2 } + b ^ { 2 } }$$

$$\textcolor{green}{ \boldsymbol{ \frac { \ln b – \ln a } { b – a } > \frac { 2 a } { a ^ { 2 } + b ^ { 2 } } } }$$

### 证明方法二：构造函数

$$\frac { \textcolor{springgreen}{\ln x} – \ln a } { \textcolor{springgreen}{x} – a } > \frac { 2 a } { a ^ { 2 } + \textcolor{springgreen}{x ^ { 2 } } } \tag{1}$$

$$f ( x ) = \left( \textcolor{springgreen}{x ^ { 2 } } + a ^ { 2 } \right) ( \textcolor{springgreen}{ \ln x } – \ln a ) – 2 a ( \textcolor{springgreen}{x} – a ) \tag{2}$$

$$f(x) > 0$$

\begin{aligned} f ^ { \prime } ( x ) \\ \\ & = 2 x ( \ln x – \ln a ) + \left( x ^ { 2 } + a ^ { 2 } \right) \frac { 1 } { x } – 2 a \\ \\ & = 2 x ( \ln x – \ln a ) + \frac { ( x – a ) ^ { 2 } } { x } \end{aligned}

$$f^{\prime} (x) > 0$$

$$f ( a ) = 0$$

$$f ( x ) > f ( a ) = 0$$

$$\left( x ^ { 2 } + a ^ { 2 } \right) ( \ln x – \ln a ) – 2 a ( x – a ) > 0$$

$$\left( b ^ { 2 } + a ^ { 2 } \right) ( \ln b – \ln a ) – 2 a ( b – a ) > 0 \Rightarrow$$

$$\left( b ^ { 2 } + a ^ { 2 } \right) ( \ln b – \ln a ) > 2 a ( b – a ) \Rightarrow$$

$$\textcolor{green}{ \boldsymbol{ \frac { \ln b – \ln a } { b – a } > \frac { 2 a } { a ^ { 2 } + b ^ { 2 } } } }$$