Question Number 161295 by Ari last updated on 15/Dec/21
$${prove}\:{that}:{x}^{\mathrm{8}} +{x}^{\mathrm{6}} −{x}^{\mathrm{3}} −{x}+\mathrm{1}>\mathrm{0},{x}\in{R} \\ $$
Commented by Ari last updated on 15/Dec/21
thanks Mr!
Commented by mr W last updated on 15/Dec/21
$$={x}^{\mathrm{8}} −{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{4}}+{x}^{\mathrm{6}} −{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{4}}+{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}+{x}^{\mathrm{2}} −{x}+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$=\left({x}^{\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$>\mathrm{0} \\ $$