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x-gt-0-y-gt-0-prove-that-xy-x-y-lt-x-




Question Number 84532 by john santu last updated on 14/Mar/20
x> 0 , y > 0 prove that  ((xy)/(x+y)) < x
$$\mathrm{x}>\:\mathrm{0}\:,\:\mathrm{y}\:>\:\mathrm{0}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{xy}}{\mathrm{x}+\mathrm{y}}\:<\:\mathrm{x} \\ $$
Commented by mr W last updated on 14/Mar/20
⇒y<x+y  ⇒(y/(x+y))<1  ⇒((xy)/(x+y))<x
$$\Rightarrow{y}<{x}+{y} \\ $$$$\Rightarrow\frac{{y}}{{x}+{y}}<\mathrm{1} \\ $$$$\Rightarrow\frac{{xy}}{{x}+{y}}<{x} \\ $$
Answered by TANMAY PANACEA last updated on 14/Mar/20
((xy)/(x+y))−x  ((xy−x^2 −xy)/(x+y))  ((−x^2 )/(x+y))<0    so ((xy)/(x+y))<x
$$\frac{{xy}}{{x}+{y}}−{x} \\ $$$$\frac{{xy}−{x}^{\mathrm{2}} −{xy}}{{x}+{y}} \\ $$$$\frac{−{x}^{\mathrm{2}} }{{x}+{y}}<\mathrm{0}\:\:\:\:{so}\:\frac{{xy}}{{x}+{y}}<{x} \\ $$

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