设 a , b > 0 a,b>0 a,b>0则
a b < a − b ln a − ln b < a + b 2 \sqrt{ab} < \frac{a - b}{\ln a - \ln b} < \frac{a + b}{2} ab<lna−lnba−b<2a+b
证明：
不妨假设 a > b > 0 a>b > 0 a>b>0
右边
a − b ln a − ln b < a + b 2 ln a − ln b > 2 ( a − b ) a + b ln a b > 2 ( a b − 1 ) a b + 1 ( a b + 1 ) ln a b − 2 ( a b − 1 ) > 0 \begin{aligned} \frac{a - b}{\ln a - \ln b} &< \frac{a + b}{2}\\ \ln a - \ln b & >\frac{2\left(a - b\right)}{a + b}\\ \ln \frac{a}{b} &> \frac{2\left(\frac{a}{b} - 1\right)}{\frac{a}{b} + 1}\\ \left(\frac{a}{b} + 1\right)\ln\frac{a}{b} - 2\left(\frac{a}{b}- 1\right) &>0 \end{aligned} lna−lnba−blna−lnblnba(ba+1)lnba−2(ba−1)<2a+b>a+b2(a−b)>ba+12(ba−1)>0
令 t = a b > 1 t = \frac{a}{b} > 1 t=ba>1
f ( t ) = ( t + 1 ) ln t − 2 ( t − 1 ) f\left(t\right) =\left(t+1\right)\ln t -2\left(t-1\right) f(t)=(t+1)lnt−2(t−1)
f ′ ( t ) = ln t + 1 t − 1 f^{\prime} \left(t\right)=\ln t +\frac{1}{t}-1 f′(t)=lnt+t1−1
f ′ ′ ( t ) = t − 1 t 2 > 0 f^{\prime\prime}\left(t\right) = \frac{t - 1}{t^2} >0 f′′(t)=t2t−1>0
f ′ ( t ) > f ′ ( 1 ) = 0 f^{\prime}\left(t\right) > f^{\prime}\left(1\right)=0 f′(t)>f′(1)=0
f ( t ) > f ( 1 ) = 0 f\left(t\right) > f\left(1\right)=0 f(t)>f(1)=0
因此成立
左边
a b < a − b ln a − ln b ln a − ln b < a b − b a 2 ln a b − a b + b a < 0 \begin{aligned} \sqrt{ab}&< \frac{a - b}{\ln a - \ln b} \\ \ln a - \ln b & <\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\\ 2\ln\sqrt{\frac{a}{b}}-\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}} &<0 \end{aligned} ablna−lnb2lnba−ba+ab<lna−lnba−b<ba−ab<0
令 t = a b > 1 t = \frac{a}{b} > 1 t=ba>1
g ( t ) = 2 ln t − t + 1 t g\left(t\right) = 2\ln t - t + \frac{1}{t} g(t)=2lnt−t+t1
g ′ ( t ) = 2 t − 1 − 1 t 2 = − ( t − 1 ) 2 t 2 < 0 g^{\prime}\left(t\right)=\frac{2}{t}-1-\frac{1}{t^2}=-\frac{\left(t-1\right)^2}{t^2}<0 g′(t)=t2−1−t21=−t2(t−1)2<0
g ( t ) < g ( 1 ) = 0 g\left(t\right) < g\left(1\right)=0 g(t)<g(1)=0
因此成立
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