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Question Number 84915 by M±th+et£s last updated on 17/Mar/20
if   x>0,y>0,z>0  show that  ((x+y)/z)+((z+y)/( x))+((z+x)/y)≥6
$${if}\: \\ $$$${x}>\mathrm{0},{y}>\mathrm{0},{z}>\mathrm{0} \\ $$$${show}\:{that} \\ $$$$\frac{{x}+{y}}{{z}}+\frac{{z}+{y}}{\:{x}}+\frac{{z}+{x}}{{y}}\geqslant\mathrm{6}\:\: \\ $$
Answered by TANMAY PANACEA last updated on 17/Mar/20
((x+y)/z)+1+((z+y)/x)+1+((z+x)/y)+1−3  (x+y+z)((1/x)+(1/y)+(1/z))−3  using AM −GM− HM  AM≥GM≥HM  AM=((x+y+z)/3)  HM=(3/((1/x)+(1/y)+(1/z)))  ((x+y+z)/3)≥(3/((1/x)+(1/y)+(1/z)))  (x+y+z)((1/x)+(1/y)+(1/z))≥9  (x+y+z)((1/x)+(1/y)+(1/z))−3≥9−3  so  (x+y+z)((1/x)+(1/y)+(1/z))≥6
$$\frac{{x}+{y}}{{z}}+\mathrm{1}+\frac{{z}+{y}}{{x}}+\mathrm{1}+\frac{{z}+{x}}{{y}}+\mathrm{1}−\mathrm{3} \\ $$$$\left({x}+{y}+{z}\right)\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right)−\mathrm{3} \\ $$$${using}\:{AM}\:−{GM}−\:{HM} \\ $$$${AM}\geqslant{GM}\geqslant{HM} \\ $$$${AM}=\frac{{x}+{y}+{z}}{\mathrm{3}} \\ $$$${HM}=\frac{\mathrm{3}}{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}} \\ $$$$\frac{{x}+{y}+{z}}{\mathrm{3}}\geqslant\frac{\mathrm{3}}{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}} \\ $$$$\left({x}+{y}+{z}\right)\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right)\geqslant\mathrm{9} \\ $$$$\left({x}+{y}+{z}\right)\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right)−\mathrm{3}\geqslant\mathrm{9}−\mathrm{3} \\ $$$${so} \\ $$$$\left({x}+{y}+{z}\right)\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right)\geqslant\mathrm{6} \\ $$
Commented by M±th+et£s last updated on 17/Mar/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by TANMAY PANACEA last updated on 17/Mar/20
mostwelcome sir
$${mostwelcome}\:{sir} \\ $$

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