Question Number 51721 by Tawa1 last updated on 29/Dec/18
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\mathrm{given}\:\mathrm{that}:\:\:\:\:\:\mid\mathrm{x}\:+\:\mathrm{i}\mid\:\geqslant\:\frac{\mathrm{1}}{\mathrm{x}} \\ $$
Answered by ajfour last updated on 29/Dec/18
$$\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\geqslant\:\frac{\mathrm{1}}{{x}} \\ $$$$\:{x}\:<\:\mathrm{0}\:\:{and}\:{if}\:{x}>\mathrm{0}\:{then} \\ $$$$\:{let}\:{x}^{\mathrm{2}} \:=\:{y} \\ $$$$\Rightarrow\:{y}\left(\mathrm{1}+{y}\right)\geqslant\mathrm{1} \\ $$$$\Rightarrow\:{y}^{\mathrm{2}} +{y}−\mathrm{1}\:\geqslant\:\mathrm{0} \\ $$$$\:\:\:\left({y}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \geqslant\:\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\Rightarrow\:{y}\:\geqslant\:\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:{or}\:\:{x}\:\geqslant\:\sqrt{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}} \\ $$$${so}\:\:\:{x}\:\in\:\left(−\infty,\:\mathrm{0}\right)\cup\left[\sqrt{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}}\:,\infty\right)\:. \\ $$
Commented by Tawa1 last updated on 30/Dec/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$