Question Number 99645 by bramlex last updated on 22/Jun/20
$${If}\:{x},{y}\:>\:\mathrm{0}\:{then}\:\mid\sqrt{{xy}}−\frac{{x}+{y}}{\mathrm{2}}\mid\:+\:\mid\frac{{x}+{y}}{\mathrm{2}}\:+\:\sqrt{{xy}}\:\mid\:= \\ $$
Commented by bemath last updated on 22/Jun/20
$$\mathrm{since}\:\sqrt{\mathrm{xy}}\:>\:\mathrm{0}\:\&\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\:>\:\mathrm{0} \\ $$$$\mathrm{so}\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\:+\:\sqrt{\mathrm{xy}}\:>\:\mathrm{0} \\ $$$$\mathrm{case}\left(\mathrm{1}\right)\:\mathrm{if}\:\sqrt{\mathrm{xy}}−\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\:>\:\mathrm{0} \\ $$$$\mathrm{then}\:\sqrt{\mathrm{xy}}\:−\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\:+\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\:+\sqrt{\mathrm{xy}}\:=\:\mathrm{2}\sqrt{\mathrm{xy}\:} \\ $$$$\mathrm{case}\left(\mathrm{2}\right)\:\mathrm{if}\:\sqrt{\mathrm{xy}}−\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\:<\:\mathrm{0} \\ $$$$\mathrm{then}\:−\sqrt{\mathrm{xy}}+\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\:+\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}+\sqrt{\mathrm{xy}}\:=\:\mathrm{x}+\mathrm{y}\: \\ $$$$ \\ $$
Answered by Farruxjano last updated on 22/Jun/20
$$\boldsymbol{{When}}\:\boldsymbol{{x}},\boldsymbol{{y}}>\mathrm{0}\:\Rightarrow\:\frac{{x}+{y}}{\mathrm{2}}\:+\:\sqrt{{xy}}>\mathrm{0}\:\:\Rightarrow \\ $$$$\mid\frac{{x}+{y}}{\mathrm{2}}\:+\:\sqrt{{xy}}\:\mid=\frac{{x}+{y}}{\mathrm{2}}\:+\:\sqrt{{xy}} \\ $$$$\boldsymbol{{We}}\:\boldsymbol{{know}}\:\boldsymbol{{inequality}}\:\boldsymbol{{Koshi}}: \\ $$$$\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{2}}\geqslant\sqrt{\boldsymbol{{xy}}}\:\:\:\boldsymbol{{when}}\:\:\boldsymbol{{x}},\boldsymbol{{y}}>\mathrm{0}\:\:\boldsymbol{{So}}, \\ $$$$\mid\sqrt{{xy}}−\frac{{x}+{y}}{\mathrm{2}}\mid=\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{2}}−\sqrt{\boldsymbol{{xy}}}\:\: \\ $$$$\mid\sqrt{{xy}}−\frac{{x}+{y}}{\mathrm{2}}\mid\:+\:\mid\frac{{x}+{y}}{\mathrm{2}}\:+\:\sqrt{{xy}}\:\mid\:=\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{2}}−\sqrt{\boldsymbol{{xy}}}+ \\ $$$$+\frac{{x}+{y}}{\mathrm{2}}\:+\:\sqrt{{xy}}=\boldsymbol{{x}}+\boldsymbol{{y}}\:\blacktriangle. \\ $$