AtomAlpaca
fmin(x)f_{min}(x) 代表函数 f(x)f(x) 的最小值；
fmax(x)f_{max}(x) 代表函数 f(x)f(x) 的最大值；
Theorem 1. 求证：(x+1)⋅1−x(lnx−1)ex<ex+ex−2(x+1) \cdot \frac{1-x(\ln x - 1)}{e^x} < e^x + e^{x-2}
Proof. 1−xlnx−x<2(x+1)x+1<ex+1+ex−1x+1⇔x+xlnx+1>0\begin{aligned} & 1 - x\ln x - x < \frac{2(x+1)}{x+1} < \frac{e^{x+1} + e^{x - 1}}{x + 1} \\ \Leftrightarrow& x + x \ln x +1 > 0 \end{aligned} 令 f(x)=x+xlnx+1f(x) = x + x \ln x +1，f′(x)=lnx+2f'(x) = \ln x+2
fmin(x)=f(e−2)=1−e−2>0f_{min}(x) = f(e^{-2}) = 1 - e^{-2} > 0，故 f(x)>0f(x) > 0 ◻
Theorem 2. f(x)=2ln(x+1)−a(x+1)2f(x) = 2\ln(x + 1) - \frac{a}{(x + 1)^2} 有两个不同零点 x1,x2x_1, x_2
当 −1<x<−12-1 < x < -\frac{1}{2} 时，求证：2ln(x+1)>−1x+12 \ln(x+1) > - \frac{1}{x+1}
求 aa 的取值范围
Proof. 1. 令 t=x+1t = x + 1，即证 t∈(o,12)t \in (o, \frac{1}{2}) 时有 2lnt>−1t2 \ln t > -\frac{1}{t} 令 g(t)=2lnt+1tg(t) = 2 \ln t + \frac{1}{t}，g′(t)=2t−1t2g'(t) = \frac{2t - 1}{t^2} gmin(t)=g(12)=2−2ln2>0g_{\min}(t) = g(\frac{1}{2}) = 2 - 2 \ln 2 > 0 ◻
2. 令 t=x+1t = x + 1，f′(t)=2t+2at2=2(t2+a)t3f'(t) = \frac{2}{t} + \frac{2a}{t^2} = \frac{2(t^2 + a)}{t^3}
a≥0a \ge 0，f(t)f(t) 单增，无解
a<0a < 0，fmin(t)=f(−a)<0⇔2ln−a+1<0f_min(t) = f(\sqrt{-a}) < 0 \Leftrightarrow 2ln \sqrt{-a} + 1 < 0 a∈(−e−1,0)a \in (-e^{-1}, 0)