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](https://www.tinkutara.com/question/Q138648.png)
$$\left.\mathrm{f}\left.\mathrm{rom}\:\mathrm{1}\right)\:\Rightarrow\:\mathrm{2}\right) \\ $$$${x}^{\mathrm{2}} \pm{xy}+{y}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +{x}^{\mathrm{2}} \pm\mathrm{2}{xy}+{y}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\left({x}\pm{y}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\geqslant\mathrm{0}\:\left[\mathrm{obviously}\right] \\ $$$$\left.\mathrm{3}\right) \\ $$$${x}^{\mathrm{2}} \pm{xy}+{y}^{\mathrm{2}} =\mathrm{0}\:\Leftrightarrow\:{y}=\left(\pm\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){x}\vee{y}=\left(\pm\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){x} \\ $$$$\Rightarrow\:\mathrm{if}\:{A}=\mathrm{0}\:\Leftrightarrow\:{x}={y}=\mathrm{0} \\ $$$$\left.\mathrm{4}\right)\:\mathrm{seems}\:\mathrm{obvious}: \\ $$$${f}\left({r}\right)={r}^{\mathrm{3}} \:\Rightarrow\:{f}'\left({r}\right)=\mathrm{3}{r}^{\mathrm{2}} \geqslant\mathrm{0}\:\forall{r}\in\mathbb{R} \\ $$$$\Rightarrow\:{f}\left({r}\right)={r}^{\mathrm{3}} \:\mathrm{is}\:\mathrm{monotonically}\:\mathrm{increasing}\:\mathrm{on}\:\mathbb{R} \\ $$$$\Rightarrow\:{x}<{y}\:\Leftrightarrow\:{x}^{\mathrm{3}} <{y}^{\mathrm{3}} \\ $$