Question and Answers Forum

All Questions      Topic List

Number Theory Questions

Previous in All Question      Next in All Question      

Previous in Number Theory      Next in Number Theory      

Question Number 112998 by Aina Samuel Temidayo last updated on 10/Sep/20

$$\mathrm{If}\mathrm{x}+\mathrm{y}+\mathrm{z}=1\mathrm{and}\mathrm{x},\mathrm{y},\mathrm{z}\mathrm{are}\mathrm{positive}$$$$\mathrm{real}\mathrm{numbers},\mathrm{then}\mathrm{the}\mathrm{least}\mathrm{value}\mathrm{of}$$$$(\frac{1}{\mathrm{x}}-1)(\frac{1}{\mathrm{y}}-1)(\frac{1}{\mathrm{z}}-1)\mathrm{is}$$

Answered by MJS_new last updated on 11/Sep/20

$$z=1-x-y$$$$\Rightarrow -\frac{(x-1)(x+y)(y-1)}{x(x+y-1)y}$$$$\frac{d}{dy}[-\frac{(x-1)(x+y)(y-1)}{x(x+y-1)y}]=0$$$$-\frac{(x-1)(x+2y-1)}{{(x+y-1)}^2{y}^2}=0\Rightarrow y=\frac{1-x}{2}$$$$\Rightarrow -\frac{{(x+1)}^2}{x(x-1)}$$$$\frac{d}{dx}[-\frac{{(x+1)}^2}{x(x-1)}]=0$$$$\frac{(x+1)(3x-1)}{x^2{(x-1)}^2}=0\Rightarrow x=y=z=\frac{1}{3}$$$$\Rightarrow \mathrm{minimum}\mathrm{value}\mathrm{is}8$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com