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Question Number 156328 by MathSh last updated on 10/Oct/21

Commented by MJS_new last updated on 10/Oct/21

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{wrong}.\mathrm{it}\mathrm{must}\mathrm{be}>2$$

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

$$\mathrm{Thank}\mathrm{You}\mathrm{Dear}\mathrm{Ser}$$$$\mathrm{Is}\mathrm{if}\mathrm{possible}\mathrm{to}\mathrm{prove}\mathrm{for}\mathrm{the}\mathrm{case}>2$$

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

$$\mathrm{we}\mathrm{know}$$$${\phi }^2=\phi +1\iff \frac{1}{\phi }=\phi -1\iff {\phi }^3={\phi }^2+\phi =2\phi +1\mathrm{etc}.$$$$\Rightarrow \mathrm{we}\mathrm{can}\mathrm{transform}$$$$\frac{\phi \sqrt{\phi }+1}{\sqrt{\phi }(1+\sqrt{\phi })}=\phi \sqrt{\phi }-1$$$$\frac{\phi }{\phi +1}=\phi -1$$$$\frac{2}{\sqrt{\phi }(\phi +\sqrt{\phi }+1)}=\phi +\sqrt{\phi }(\phi -1)-2$$$$\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{these}\mathrm{is}$$$$\phi \sqrt{\phi }-1+\phi -1+\phi +\sqrt{\phi }(\phi -1)-2=$$$$=2\phi -4+(2\phi -1)\sqrt{\phi }=$$$$[2\phi -1=\sqrt{5}]$$$$=\sqrt{5}-3+\sqrt{5\phi }$$$$\mathrm{now}\mathrm{we}\mathrm{prove}\mathrm{this}\mathrm{is}\mathrm{greater}\mathrm{than}2$$$$\sqrt{5}-3+\sqrt{5\phi }>2\iff \sqrt{\phi }>\sqrt{5}-1\iff \phi >6-2\sqrt{5}$$$$\frac{1}{2}+\frac{\sqrt{5}}{2}>6-2\sqrt{5}$$$$1+\sqrt{5}>12-4\sqrt{5}$$$$\sqrt{5}>\frac{11}{5}$$$$5>\frac{121}{25}$$$$\frac{125}{25}>\frac{121}{25}$$$$\mathrm{true}$$

Commented by MathSh last updated on 10/Oct/21

$$\mathrm{Perfect}\mathrm{my}\mathrm{dear}\mathbf{S}\mathrm{er},\mathrm{thank}\mathrm{you}$$

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