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

Show that: y + (√(y^2 − 1)) ≥ 1 and 0 < y − (√(y^2 − 1)) ≤ 1 if y ≥ 1

$$\mathrm{Show}\:\mathrm{that}:\:\:\:\:\:\:\mathrm{y}\:\:+\:\:\sqrt{\mathrm{y}^{\mathrm{2}} \:−\:\mathrm{1}}\:\:\:\geqslant\:\:\mathrm{1}\:\:\:\:\:\mathrm{and}\:\:\:\:\mathrm{0}\:\:<\:\:\mathrm{y}\:\:−\:\:\sqrt{\mathrm{y}^{\mathrm{2}} \:−\:\mathrm{1}}\:\:\leqslant\:\:\mathrm{1} \\ $$ $$\mathrm{if}\:\:\mathrm{y}\:\:\geqslant\:\mathrm{1} \\ $$

Answered by MJS last updated on 26/Feb/20

t^2 =y^2 −1 y=±(√(t^2 +1))∧y≥1 ⇒ y=(√(t^2 +1)) y+(√(y^2 −1))≥1 (√(t^2 +1))+∣t∣≥1 minimum of (√(t^2 +1)) is 1 at t=0 minimum of ∣t∣ is 0 at t=0 ⇒ (√(t^2 +1))+∣t∣≥1 0<y−(√(y^2 −1))≤1 0<(√(t^2 +1))−(√t^2 )≤1 (√t^2 )<(√(t^2 +1))≤(√t^2 )+1 (√t^2 )<(√(t^2 +1)) all sides >0 ⇒ we are allowed to square t^2 <t^2 +1≤t^2 +1+2∣t∣ always true

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

Commented byTawaTawa1 last updated on 26/Feb/20

God bless you sir. I appreciate.

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

Commented byTawaTawa1 last updated on 26/Feb/20

Sir, please next one. Q82953.

$$\mathrm{Sir},\:\mathrm{please}\:\:\mathrm{next}\:\mathrm{one}.\:\:\:\mathrm{Q82953}. \\ $$

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