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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 byTawaTawa1 last updated on 26/Feb/20

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

Commented byTawaTawa1 last updated on 26/Feb/20

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

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