Question Number 80816 by ~blr237~ last updated on 06/Feb/20
$${let}\:\:\mathrm{0}<{a}<{b}\:\:{prove}\:{that} \\ $$$$\:{ln}\left(\mathrm{1}+\frac{{a}}{{b}}\right){ln}\left(\mathrm{1}+\frac{{b}}{{a}}\right)<\:\left({ln}\mathrm{2}\right)^{\mathrm{2}} \:\: \\ $$
Commented by ~blr237~ last updated on 06/Feb/20
$${Sir}\:\:,\:{i}\:{look}\:{like}\:\:{at}\:\left(\mathrm{3}−\mathrm{4}\right)\:{crossing}\:{you}\:{uze} \\ $$$${A}\leqslant{B}\:{and}\:{B}\geqslant{C}\:\Rightarrow\:{A}\leqslant{C}\:\:\:\:??\: \\ $$$${if}\:{not}\:,{please}\:{explain} \\ $$
Commented by mind is power last updated on 07/Feb/20
$${mistack}\:{sorry} \\ $$
Answered by mr W last updated on 07/Feb/20
![let x=(a/b)≠1 f(x)=ln (1+x)×ln (1+(1/x)) f′(x)=(1/(1+x))ln (1+(1/x))−(1/x^2 )×(1/(1+(1/x)))ln (1+x) f′(x)=(1/(x(1+x)))[xln (1+(1/x))−ln (1+x)]=0 xln (1+(1/x))−ln (1+x)=0 ln (1+(1/x))^x =ln (1+x) (1+(1/x))^x =(1+x) ⇒x=1 f_(max) =f(1)=(ln 2)^2 ⇒f(x)≤(ln 2)^2 for x≠1: f(x)<(ln 2)^2](https://www.tinkutara.com/question/Q80856.png)
$${let}\:{x}=\frac{{a}}{{b}}\neq\mathrm{1} \\ $$$${f}\left({x}\right)=\mathrm{ln}\:\left(\mathrm{1}+{x}\right)×\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right) \\ $$$${f}'\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}}\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{x}}}\mathrm{ln}\:\left(\mathrm{1}+{x}\right) \\ $$$${f}'\left({x}\right)=\frac{\mathrm{1}}{{x}\left(\mathrm{1}+{x}\right)}\left[{x}\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)−\mathrm{ln}\:\left(\mathrm{1}+{x}\right)\right]=\mathrm{0} \\ $$$${x}\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)−\mathrm{ln}\:\left(\mathrm{1}+{x}\right)=\mathrm{0} \\ $$$$\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} =\mathrm{ln}\:\left(\mathrm{1}+{x}\right) \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} =\left(\mathrm{1}+{x}\right) \\ $$$$\Rightarrow{x}=\mathrm{1} \\ $$$${f}_{{max}} ={f}\left(\mathrm{1}\right)=\left(\mathrm{ln}\:\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)\leqslant\left(\mathrm{ln}\:\mathrm{2}\right)^{\mathrm{2}} \\ $$$${for}\:{x}\neq\mathrm{1}: \\ $$$${f}\left({x}\right)<\left(\mathrm{ln}\:\mathrm{2}\right)^{\mathrm{2}} \\ $$