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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 by TawaTawa1 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 by TawaTawa1 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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