Question Number 108051 by ZiYangLee last updated on 14/Aug/20
Answered by Sarah85 last updated on 14/Aug/20
$$\sqrt{{x}}−\sqrt{{y}}\geqslant\mathrm{0} \\ $$$$\left(\sqrt{{x}}−\sqrt{{y}}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${x}−\mathrm{2}\sqrt{{xy}}+{y}\geqslant\mathrm{0} \\ $$$${x}+{y}\geqslant\mathrm{2}\sqrt{{xy}} \\ $$$$\frac{{x}+{y}}{\mathrm{2}}\geqslant\sqrt{{xy}} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{rectangle} \\ $$$$\mathrm{perimeter}\:\mathrm{2}{x}+\mathrm{2}{y} \\ $$$$\mathrm{area}\:{xy} \\ $$$$ \\ $$$$\mathrm{square} \\ $$$$\mathrm{perimeter}\:\mathrm{2}{x}+\mathrm{2}{y}=\mathrm{4}{s}\:\Leftrightarrow\:{s}=\frac{{x}+{y}}{\mathrm{2}} \\ $$$$\mathrm{area}\:{s}^{\mathrm{2}} =\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{showed} \\ $$$$\frac{{x}+{y}}{\mathrm{2}}\geqslant\sqrt{{xy}};\:{x}>\mathrm{0}\wedge{y}>\mathrm{0}\:\Rightarrow \\ $$$$\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{\mathrm{2}} \geqslant{xy} \\ $$$$\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{\mathrm{2}} ={xy}\:\Leftrightarrow\:{x}={y} \\ $$