Question Number 156328 by MathSh last updated on 10/Oct/21
Commented by MJS_new last updated on 10/Oct/21
$$\mathrm{it}'\mathrm{s}\:\mathrm{wrong}.\:\mathrm{it}\:\mathrm{must}\:\mathrm{be}\:>\mathrm{2} \\ $$
Commented by MathSh last updated on 10/Oct/21
$$\mathrm{Good}\:\mathrm{dear}\:\mathrm{Ser}\:\mathrm{thankyou} \\ $$
Commented by MathSh last updated on 10/Oct/21
$$\mathrm{Thank}\:\mathrm{You}\:\mathrm{Dear}\:\mathrm{Ser} \\ $$$$\mathrm{Is}\:\mathrm{if}\:\:\mathrm{possible}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{for}\:\mathrm{the}\:\mathrm{case}\:>\mathrm{2} \\ $$
Commented by MJS_new last updated on 10/Oct/21
$$\mathrm{I}\:\mathrm{will}\:\mathrm{post}\:\mathrm{it}\:\mathrm{later} \\ $$
Answered by MJS_new last updated on 10/Oct/21
![we know ϕ^2 =ϕ+1 ⇔ (1/ϕ)=ϕ−1 ⇔ ϕ^3 =ϕ^2 +ϕ=2ϕ+1 etc. ⇒ we can transform ((ϕ(√ϕ)+1)/( (√ϕ)(1+(√ϕ))))=ϕ(√ϕ)−1 (ϕ/(ϕ+1))=ϕ−1 (2/( (√ϕ)(ϕ+(√ϕ)+1)))=ϕ+(√ϕ)(ϕ−1)−2 the sum of these is ϕ(√ϕ)−1+ϕ−1+ϕ+(√ϕ)(ϕ−1)−2= =2ϕ−4+(2ϕ−1)(√ϕ)= [2ϕ−1=(√5)] =(√5)−3+(√(5ϕ)) now we prove this is greater than 2 (√5)−3+(√(5ϕ))>2 ⇔ (√ϕ)>(√5)−1 ⇔ ϕ>6−2(√5) (1/2)+((√5)/2)>6−2(√5) 1+(√5)>12−4(√5) (√5)>((11)/5) 5>((121)/(25)) ((125)/(25))>((121)/(25)) true](https://www.tinkutara.com/question/Q156408.png)
$$\mathrm{we}\:\mathrm{know} \\ $$$$\varphi^{\mathrm{2}} =\varphi+\mathrm{1}\:\Leftrightarrow\:\frac{\mathrm{1}}{\varphi}=\varphi−\mathrm{1}\:\Leftrightarrow\:\varphi^{\mathrm{3}} =\varphi^{\mathrm{2}} +\varphi=\mathrm{2}\varphi+\mathrm{1}\:\mathrm{etc}. \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{can}\:\mathrm{transform} \\ $$$$\frac{\varphi\sqrt{\varphi}+\mathrm{1}}{\:\sqrt{\varphi}\left(\mathrm{1}+\sqrt{\varphi}\right)}=\varphi\sqrt{\varphi}−\mathrm{1} \\ $$$$\frac{\varphi}{\varphi+\mathrm{1}}=\varphi−\mathrm{1} \\ $$$$\frac{\mathrm{2}}{\:\sqrt{\varphi}\left(\varphi+\sqrt{\varphi}+\mathrm{1}\right)}=\varphi+\sqrt{\varphi}\left(\varphi−\mathrm{1}\right)−\mathrm{2} \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{these}\:\mathrm{is} \\ $$$$\varphi\sqrt{\varphi}−\mathrm{1}+\varphi−\mathrm{1}+\varphi+\sqrt{\varphi}\left(\varphi−\mathrm{1}\right)−\mathrm{2}= \\ $$$$=\mathrm{2}\varphi−\mathrm{4}+\left(\mathrm{2}\varphi−\mathrm{1}\right)\sqrt{\varphi}= \\ $$$$\:\:\:\:\:\left[\mathrm{2}\varphi−\mathrm{1}=\sqrt{\mathrm{5}}\right] \\ $$$$=\sqrt{\mathrm{5}}−\mathrm{3}+\sqrt{\mathrm{5}\varphi} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{prove}\:\mathrm{this}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{2} \\ $$$$\sqrt{\mathrm{5}}−\mathrm{3}+\sqrt{\mathrm{5}\varphi}>\mathrm{2}\:\Leftrightarrow\:\sqrt{\varphi}>\sqrt{\mathrm{5}}−\mathrm{1}\:\Leftrightarrow\:\varphi>\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}>\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$\mathrm{1}+\sqrt{\mathrm{5}}>\mathrm{12}−\mathrm{4}\sqrt{\mathrm{5}} \\ $$$$\sqrt{\mathrm{5}}>\frac{\mathrm{11}}{\mathrm{5}} \\ $$$$\mathrm{5}>\frac{\mathrm{121}}{\mathrm{25}} \\ $$$$\frac{\mathrm{125}}{\mathrm{25}}>\frac{\mathrm{121}}{\mathrm{25}} \\ $$$$\mathrm{true} \\ $$
Commented by MathSh last updated on 10/Oct/21
$$\mathrm{Perfect}\:\mathrm{my}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you} \\ $$