Question Number 57585 by Mikael_Marshall last updated on 07/Apr/19
$${knowing}\:{that}\:{x}+{y}=\mathrm{1}.\:{what}\:{is}\:{the}\:{result}\:{of}\:\frac{{y}}{{x}}+\frac{{x}}{{y}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 08/Apr/19
$$\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{xy}} \\ $$$${R}=\frac{\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}}{{xy}} \\ $$$${R}=\frac{\left({x}+{y}\right)^{\mathrm{2}} }{{xy}}−\mathrm{2} \\ $$$$\frac{{x}+{y}}{\mathrm{2}}\geqslant\sqrt{{xy}} \\ $$$$\frac{\left({x}+{y}\right)^{\mathrm{2}} }{{xy}}\geqslant\mathrm{4} \\ $$$${R}\geqslant\mathrm{4}−\mathrm{2} \\ $$$${R}\geqslant\mathrm{2} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 08/Apr/19
$${ok}\:{sir}... \\ $$
Commented by MJS last updated on 08/Apr/19
$$\frac{{x}+{y}}{\mathrm{2}}\geqslant\sqrt{{xy}}\:\mathrm{only}\:\mathrm{makes}\:\mathrm{sense}\:\mathrm{for}\:{x}\geqslant\mathrm{0}\:\wedge\:{y}\geqslant\mathrm{0} \\ $$
Answered by MJS last updated on 07/Apr/19
$${y}=\mathrm{1}−{x} \\ $$$$\frac{\mathrm{1}−{x}}{{x}}+\frac{{x}}{\mathrm{1}−{x}}=\frac{\mathrm{1}}{{x}\left(\mathrm{1}−{x}\right)}−\mathrm{2} \\ $$