Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 138641 by soudo last updated on 15/Apr/21

Answered by MJS_new last updated on 16/Apr/21

from 1) ⇒ 2)  x^2 ±xy+y^2 =(1/2)(x^2 +x^2 ±2xy+y^2 +y^2 )=  =(1/2)((x±y)^2 +x^2 +y^2 )≥0 [obviously]  3)  x^2 ±xy+y^2 =0 ⇔ y=(±(1/2)−((√3)/2)i)x∨y=(±(1/2)+((√3)/2)i)x  ⇒ if A=0 ⇔ x=y=0  4) seems obvious:  f(r)=r^3  ⇒ f′(r)=3r^2 ≥0 ∀r∈R  ⇒ f(r)=r^3  is monotonically increasing on R  ⇒ x<y ⇔ x^3 <y^3

$$\left.\mathrm{f}\left.\mathrm{rom}\:\mathrm{1}\right)\:\Rightarrow\:\mathrm{2}\right) \\ $$$${x}^{\mathrm{2}} \pm{xy}+{y}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +{x}^{\mathrm{2}} \pm\mathrm{2}{xy}+{y}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\left({x}\pm{y}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\geqslant\mathrm{0}\:\left[\mathrm{obviously}\right] \\ $$$$\left.\mathrm{3}\right) \\ $$$${x}^{\mathrm{2}} \pm{xy}+{y}^{\mathrm{2}} =\mathrm{0}\:\Leftrightarrow\:{y}=\left(\pm\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){x}\vee{y}=\left(\pm\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){x} \\ $$$$\Rightarrow\:\mathrm{if}\:{A}=\mathrm{0}\:\Leftrightarrow\:{x}={y}=\mathrm{0} \\ $$$$\left.\mathrm{4}\right)\:\mathrm{seems}\:\mathrm{obvious}: \\ $$$${f}\left({r}\right)={r}^{\mathrm{3}} \:\Rightarrow\:{f}'\left({r}\right)=\mathrm{3}{r}^{\mathrm{2}} \geqslant\mathrm{0}\:\forall{r}\in\mathbb{R} \\ $$$$\Rightarrow\:{f}\left({r}\right)={r}^{\mathrm{3}} \:\mathrm{is}\:\mathrm{monotonically}\:\mathrm{increasing}\:\mathrm{on}\:\mathbb{R} \\ $$$$\Rightarrow\:{x}<{y}\:\Leftrightarrow\:{x}^{\mathrm{3}} <{y}^{\mathrm{3}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com