Question Number 138641 by soudo last updated on 15/Apr/21
Answered by MJS_new last updated on 16/Apr/21
$$\left.\mathrm{f}\left.\mathrm{rom}\:\mathrm{1}\right)\:\Rightarrow\:\mathrm{2}\right) \\ $$$${x}^{\mathrm{2}} \pm{xy}+{y}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +{x}^{\mathrm{2}} \pm\mathrm{2}{xy}+{y}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\left({x}\pm{y}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\geqslant\mathrm{0}\:\left[\mathrm{obviously}\right] \\ $$$$\left.\mathrm{3}\right) \\ $$$${x}^{\mathrm{2}} \pm{xy}+{y}^{\mathrm{2}} =\mathrm{0}\:\Leftrightarrow\:{y}=\left(\pm\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){x}\vee{y}=\left(\pm\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){x} \\ $$$$\Rightarrow\:\mathrm{if}\:{A}=\mathrm{0}\:\Leftrightarrow\:{x}={y}=\mathrm{0} \\ $$$$\left.\mathrm{4}\right)\:\mathrm{seems}\:\mathrm{obvious}: \\ $$$${f}\left({r}\right)={r}^{\mathrm{3}} \:\Rightarrow\:{f}'\left({r}\right)=\mathrm{3}{r}^{\mathrm{2}} \geqslant\mathrm{0}\:\forall{r}\in\mathbb{R} \\ $$$$\Rightarrow\:{f}\left({r}\right)={r}^{\mathrm{3}} \:\mathrm{is}\:\mathrm{monotonically}\:\mathrm{increasing}\:\mathrm{on}\:\mathbb{R} \\ $$$$\Rightarrow\:{x}<{y}\:\Leftrightarrow\:{x}^{\mathrm{3}} <{y}^{\mathrm{3}} \\ $$