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Question Number 56775 by zambolly19 last updated on 23/Mar/19

if x,y∍ℜ,show that ∣x+y∣=∣x∣+∣y∣ iff xy≥0

$${if}\:{x},{y}\backepsilon\Re,{show}\:{that}\:\mid{x}+{y}\mid=\mid{x}\mid+\mid{y}\mid\:{iff}\:{xy}\geqslant\mathrm{0} \\ $$

Commented by maxmathsup by imad last updated on 23/Mar/19

we have ∣x+y∣^2 −(∣x∣+∣y∣)^2  =x^2 +2xy+y^2  −x^2 −2∣xy∣−y^2   =2(xy−∣xy∣) =0 because xy≥0 ⇒∣x+y∣=∣x∣+∣y∣ .

$${we}\:{have}\:\mid{x}+{y}\mid^{\mathrm{2}} −\left(\mid{x}\mid+\mid{y}\mid\right)^{\mathrm{2}} \:={x}^{\mathrm{2}} +\mathrm{2}{xy}+{y}^{\mathrm{2}} \:−{x}^{\mathrm{2}} −\mathrm{2}\mid{xy}\mid−{y}^{\mathrm{2}} \\ $$$$=\mathrm{2}\left({xy}−\mid{xy}\mid\right)\:=\mathrm{0}\:{because}\:{xy}\geqslant\mathrm{0}\:\Rightarrow\mid{x}+{y}\mid=\mid{x}\mid+\mid{y}\mid\:. \\ $$

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