Question Number 190928 by TUN last updated on 14/Apr/23
$${Solve}\:{the}\:{equation}\:: \\ $$$$\mathrm{2}+{x}^{\mathrm{2}} \left({x}^{\mathrm{4}} +\mathrm{1}\right)=\frac{\sqrt{\mathrm{2}}{x}^{\mathrm{3}} \left({x}^{\mathrm{4}} −\mathrm{1}\right)}{\:\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}} \\ $$$$ \\ $$
Commented by Frix last updated on 14/Apr/23
$${x}=\pm\mathrm{i} \\ $$
Commented by TUN last updated on 14/Apr/23
$${it}\:{means}\:{no}\:{real}\:{root} \\ $$
Commented by TUN last updated on 14/Apr/23
$${prove}\:{it} \\ $$
Commented by Frix last updated on 14/Apr/23
$$\mathrm{2}+{x}^{\mathrm{2}} \left({x}^{\mathrm{4}} +\mathrm{1}\right)−\frac{\sqrt{\mathrm{2}}{x}^{\mathrm{3}} \left({x}^{\mathrm{4}} −\mathrm{1}\right)}{\:\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}\geqslant\mathrm{2} \\ $$
Commented by TUN last updated on 15/Apr/23
$${how}\:{to}\:\geqslant\mathrm{2} \\ $$
Commented by Frix last updated on 15/Apr/23
$$\mathrm{2}+{x}^{\mathrm{2}} \left({x}^{\mathrm{4}} +\mathrm{1}\right)−\frac{\sqrt{\mathrm{2}}{x}^{\mathrm{3}} \left({x}^{\mathrm{4}} −\mathrm{1}\right)}{\:\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}\geqslant\mathrm{2} \\ $$$${x}^{\mathrm{2}} \left({x}^{\mathrm{4}} +\mathrm{1}\right)−\frac{\sqrt{\mathrm{2}}{x}^{\mathrm{3}} \left({x}^{\mathrm{4}} −\mathrm{1}\right)}{\:\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}\geqslant\mathrm{0} \\ $$$$\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\sqrt{\mathrm{2}}{x}\left({x}^{\mathrm{4}} −\mathrm{1}\right)\geqslant\mathrm{0} \\ $$$${f}\left({x}\right)=\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\sqrt{\mathrm{2}}{x}\left({x}^{\mathrm{4}} −\mathrm{1}\right) \\ $$$${f}'\left({x}\right)=\mathrm{0}\:\Rightarrow\:{x}\approx−.\mathrm{522499270} \\ $$$${f}''\left(−.\mathrm{522499270}\right)>\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{Minimum}\:\left({f}\left({x}\right)\right)\:=.\mathrm{430}…\:>\mathrm{0} \\ $$