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Question Number 80816 by ~blr237~ last updated on 06/Feb/20

$$let0<a<bprovethat$$ $$ln(1+\frac{a}{b})ln(1+\frac{b}{a})<{\left(ln2\right)}^2$$

Commented by~blr237~ last updated on 06/Feb/20

$$Sir,ilooklikeat(3-4)crossingyouuze$$ $$A⩽BandB⩾C\Rightarrow A⩽C??$$ $$ifnot,pleaseexplain$$

Commented bymind is power last updated on 07/Feb/20

$$mistacksorry$$

Answered by mr W last updated on 07/Feb/20

$$letx=\frac{a}{b}\ne 1$$ $$f\left(x\right)=\ln (1+x)\times \ln (1+\frac{1}{x})$$ $$f^{\prime }\left(x\right)=\frac{1}{1+x}\ln (1+\frac{1}{x})-\frac{1}{x^2}\times \frac{1}{1+\frac{1}{x}}\ln (1+x)$$ $$f^{\prime }\left(x\right)=\frac{1}{x(1+x)}[x\ln (1+\frac{1}{x})-\ln (1+x)]=0$$ $$x\ln (1+\frac{1}{x})-\ln (1+x)=0$$ $$\ln {(1+\frac{1}{x})}^x=\ln (1+x)$$ $${(1+\frac{1}{x})}^x=(1+x)$$ $$\Rightarrow x=1$$ $$f_{max}=f\left(1\right)={\left(\ln 2\right)}^2$$ $$\Rightarrow f\left(x\right)⩽{\left(\ln 2\right)}^2$$ $$forx\ne 1:$$ $$f\left(x\right)<{\left(\ln 2\right)}^2$$

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