Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 82952 by TawaTawa1 last updated on 26/Feb/20

$$\mathrm{Show}\mathrm{that}:\mathrm{y}+\sqrt{{\mathrm{y}}^2-1}⩾1\mathrm{and}0<\mathrm{y}-\sqrt{{\mathrm{y}}^2-1}⩽1$$$$\mathrm{if}\mathrm{y}⩾1$$

Answered by MJS last updated on 26/Feb/20

$$t^2=y^2-1$$$$y=\pm \sqrt{t^2+1}\wedge y⩾1\Rightarrow y=\sqrt{t^2+1}$$$$y+\sqrt{y^2-1}⩾1$$$$\sqrt{t^2+1}+\mid t\mid ⩾1$$$$\mathrm{minimum}\mathrm{of}\sqrt{t^2+1}\mathrm{is}1\mathrm{at}t=0$$$$\mathrm{minimum}\mathrm{of}\mid t\mid \mathrm{is}0\mathrm{at}t=0$$$$\Rightarrow \sqrt{t^2+1}+\mid t\mid ⩾1$$$$$$$$0<y-\sqrt{y^2-1}⩽1$$$$0<\sqrt{t^2+1}-\sqrt{t^2}⩽1$$$$\sqrt{t^2}<\sqrt{t^2+1}⩽\sqrt{t^2}+1$$$$\sqrt{t^2}<\sqrt{t^2+1}$$$$\mathrm{all}\mathrm{sides}>0\Rightarrow \mathrm{we}\mathrm{are}\mathrm{allowed}\mathrm{to}\mathrm{square}$$$$t^2<t^2+1⩽t^2+1+2\mid t\mid \mathrm{always}\mathrm{true}$$

Commented by TawaTawa1 last updated on 26/Feb/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

Commented by TawaTawa1 last updated on 26/Feb/20

$$\mathrm{Sir},\mathrm{please}\mathrm{next}\mathrm{one}.\mathrm{Q}82953.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com