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Question Number 99645 by bramlex last updated on 22/Jun/20

If x,y > 0 then ∣(√(xy))−((x+y)/2)∣ + ∣((x+y)/2) + (√(xy)) ∣ =

$${If}\:{x},{y}\:>\:\mathrm{0}\:{then}\:\mid\sqrt{{xy}}−\frac{{x}+{y}}{\mathrm{2}}\mid\:+\:\mid\frac{{x}+{y}}{\mathrm{2}}\:+\:\sqrt{{xy}}\:\mid\:= \\ $$

Commented bybemath last updated on 22/Jun/20

since (√(xy)) > 0 & ((x+y)/2) > 0 so ((x+y)/2) + (√(xy)) > 0 case(1) if (√(xy))−((x+y)/2) > 0 then (√(xy)) −((x+y)/2) + ((x+y)/2) +(√(xy)) = 2(√(xy )) case(2) if (√(xy))−((x+y)/2) < 0 then −(√(xy))+((x+y)/2) + ((x+y)/2)+(√(xy)) = x+y

$$\mathrm{since}\:\sqrt{\mathrm{xy}}\:>\:\mathrm{0}\:\&\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\:>\:\mathrm{0} \\ $$ $$\mathrm{so}\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\:+\:\sqrt{\mathrm{xy}}\:>\:\mathrm{0} \\ $$ $$\mathrm{case}\left(\mathrm{1}\right)\:\mathrm{if}\:\sqrt{\mathrm{xy}}−\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\:>\:\mathrm{0} \\ $$ $$\mathrm{then}\:\sqrt{\mathrm{xy}}\:−\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\:+\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\:+\sqrt{\mathrm{xy}}\:=\:\mathrm{2}\sqrt{\mathrm{xy}\:} \\ $$ $$\mathrm{case}\left(\mathrm{2}\right)\:\mathrm{if}\:\sqrt{\mathrm{xy}}−\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\:<\:\mathrm{0} \\ $$ $$\mathrm{then}\:−\sqrt{\mathrm{xy}}+\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\:+\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}+\sqrt{\mathrm{xy}}\:=\:\mathrm{x}+\mathrm{y}\: \\ $$ $$ \\ $$

Answered by Farruxjano last updated on 22/Jun/20

When x,y>0 ⇒ ((x+y)/2) + (√(xy))>0 ⇒ ∣((x+y)/2) + (√(xy)) ∣=((x+y)/2) + (√(xy)) We know inequality Koshi: ((x+y)/2)≥(√(xy)) when x,y>0 So, ∣(√(xy))−((x+y)/2)∣=((x+y)/2)−(√(xy)) ∣(√(xy))−((x+y)/2)∣ + ∣((x+y)/2) + (√(xy)) ∣ =((x+y)/2)−(√(xy))+ +((x+y)/2) + (√(xy))=x+y ▲.

$$\boldsymbol{{When}}\:\boldsymbol{{x}},\boldsymbol{{y}}>\mathrm{0}\:\Rightarrow\:\frac{{x}+{y}}{\mathrm{2}}\:+\:\sqrt{{xy}}>\mathrm{0}\:\:\Rightarrow \\ $$ $$\mid\frac{{x}+{y}}{\mathrm{2}}\:+\:\sqrt{{xy}}\:\mid=\frac{{x}+{y}}{\mathrm{2}}\:+\:\sqrt{{xy}} \\ $$ $$\boldsymbol{{We}}\:\boldsymbol{{know}}\:\boldsymbol{{inequality}}\:\boldsymbol{{Koshi}}: \\ $$ $$\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{2}}\geqslant\sqrt{\boldsymbol{{xy}}}\:\:\:\boldsymbol{{when}}\:\:\boldsymbol{{x}},\boldsymbol{{y}}>\mathrm{0}\:\:\boldsymbol{{So}}, \\ $$ $$\mid\sqrt{{xy}}−\frac{{x}+{y}}{\mathrm{2}}\mid=\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{2}}−\sqrt{\boldsymbol{{xy}}}\:\: \\ $$ $$\mid\sqrt{{xy}}−\frac{{x}+{y}}{\mathrm{2}}\mid\:+\:\mid\frac{{x}+{y}}{\mathrm{2}}\:+\:\sqrt{{xy}}\:\mid\:=\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{2}}−\sqrt{\boldsymbol{{xy}}}+ \\ $$ $$+\frac{{x}+{y}}{\mathrm{2}}\:+\:\sqrt{{xy}}=\boldsymbol{{x}}+\boldsymbol{{y}}\:\blacktriangle. \\ $$

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