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Question Number 108051 by ZiYangLee last updated on 14/Aug/20

Answered by Sarah85 last updated on 14/Aug/20

(√x)−(√y)≥0  ((√x)−(√y))^2 ≥0  x−2(√(xy))+y≥0  x+y≥2(√(xy))  ((x+y)/2)≥(√(xy))      rectangle  perimeter 2x+2y  area xy    square  perimeter 2x+2y=4s ⇔ s=((x+y)/2)  area s^2 =(((x+y)/2))^2     we showed  ((x+y)/2)≥(√(xy)); x>0∧y>0 ⇒  (((x+y)/2))^2 ≥xy  (((x+y)/2))^2 =xy ⇔ x=y

$$\sqrt{{x}}−\sqrt{{y}}\geqslant\mathrm{0} \\ $$$$\left(\sqrt{{x}}−\sqrt{{y}}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${x}−\mathrm{2}\sqrt{{xy}}+{y}\geqslant\mathrm{0} \\ $$$${x}+{y}\geqslant\mathrm{2}\sqrt{{xy}} \\ $$$$\frac{{x}+{y}}{\mathrm{2}}\geqslant\sqrt{{xy}} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{rectangle} \\ $$$$\mathrm{perimeter}\:\mathrm{2}{x}+\mathrm{2}{y} \\ $$$$\mathrm{area}\:{xy} \\ $$$$ \\ $$$$\mathrm{square} \\ $$$$\mathrm{perimeter}\:\mathrm{2}{x}+\mathrm{2}{y}=\mathrm{4}{s}\:\Leftrightarrow\:{s}=\frac{{x}+{y}}{\mathrm{2}} \\ $$$$\mathrm{area}\:{s}^{\mathrm{2}} =\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{showed} \\ $$$$\frac{{x}+{y}}{\mathrm{2}}\geqslant\sqrt{{xy}};\:{x}>\mathrm{0}\wedge{y}>\mathrm{0}\:\Rightarrow \\ $$$$\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{\mathrm{2}} \geqslant{xy} \\ $$$$\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{\mathrm{2}} ={xy}\:\Leftrightarrow\:{x}={y} \\ $$

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