Show-that-y-y-2-1-1-and-0-lt-y-y-2-1-1-if-y-1-

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

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

if y ⩾ 1

Answered by MJS last updated on 26/Feb/20

t^{2} = y^{2} - 1

y = \pm \sqrt{t^{2} + 1} \land y ⩾ 1 \Rightarrow y = \sqrt{t^{2} + 1}

y + \sqrt{y^{2} - 1} ⩾ 1

\sqrt{t^{2} + 1} + ∣ t ∣⩾ 1

minimum of \sqrt{t^{2} + 1} is 1 at t = 0

minimum of ∣ t ∣ is 0 at t = 0

\Rightarrow \sqrt{t^{2} + 1} + ∣ t ∣⩾ 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}

all sides > 0 \Rightarrow we are allowed to square

t^{2} < t^{2} + 1 ⩽ t^{2} + 1 + 2 ∣ t ∣ always true

Commented by TawaTawa1 last updated on 26/Feb/20

God bless you sir . I appreciate .

Commented by TawaTawa1 last updated on 26/Feb/20

Sir, please next one . Q 82953 .