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Question Number 13155 by Tinkutara last updated on 15/May/17

Answered by mrW1 last updated on 16/May/17

=(((x+y)/x))(((y+z)/y))(((z+x)/z)) =(1+(y/x))(1+(z/y))(1+(x/z)) ≥2(√(y/x))×2(√(z/y))×2(√(x/z))=8 x+y+z=1 is not necessary!

$$=\left(\frac{{x}+{y}}{{x}}\right)\left(\frac{{y}+{z}}{{y}}\right)\left(\frac{{z}+{x}}{{z}}\right) \\ $$$$=\left(\mathrm{1}+\frac{{y}}{{x}}\right)\left(\mathrm{1}+\frac{{z}}{{y}}\right)\left(\mathrm{1}+\frac{{x}}{{z}}\right) \\ $$$$\geqslant\mathrm{2}\sqrt{\frac{{y}}{{x}}}×\mathrm{2}\sqrt{\frac{{z}}{{y}}}×\mathrm{2}\sqrt{\frac{{x}}{{z}}}=\mathrm{8} \\ $$$$ \\ $$$${x}+{y}+{z}=\mathrm{1}\:{is}\:{not}\:{necessary}! \\ $$

Commented by Tinkutara last updated on 16/May/17

Thanks!

$$\mathrm{Thanks}! \\ $$

Commented by Tinkutara last updated on 16/May/17

Can I contact you on WhatsApp? Or on GMail?

$$\mathrm{Can}\:\mathrm{I}\:\mathrm{contact}\:\mathrm{you}\:\mathrm{on}\:\mathrm{WhatsApp}?\:\mathrm{Or}\: \\ $$$$\mathrm{on}\:\mathrm{GMail}? \\ $$

Commented by Tinkutara last updated on 16/May/17

Sir, can you please solve Q No. 13127 on Geometry?

$$\mathrm{Sir},\:\mathrm{can}\:\mathrm{you}\:\mathrm{please}\:\mathrm{solve}\:\mathrm{Q}\:\mathrm{No}.\:\mathrm{13127}\:\mathrm{on}\:\mathrm{Geometry}? \\ $$

Commented by mrW1 last updated on 17/May/17

Ok, I′ll try.

$${Ok},\:{I}'{ll}\:{try}. \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17

((x+y)/z).((y+z)/x).((z+x)/y)=((1−z)/z).((1−x)/x).((1−y)/y)= ((1/z)−1)((1/x)−1)((1/y)−1)= ((1/(xz))−(1/z)−(1/x)+1)((1/y)−1)= =(1/(xyz))−(1/(zy))−(1/(xy))−(1/(xz))+(1/z)+(1/x)+(1/y)−1= =(1/(xyz))−((x+y+z)/(xyz))+(1/x)+(1/y)+(1/z)−1= =(1/x)+(1/y)+(1/z)−1≥3((1/(xyz)))^(1/3) −1=(3/((xyz))^(1/3) )−1≥ ≥(3/(1/3))−1=9−1=8 .■ note: x+y+z≥3((xyz))^(1/3) ⇒((xyz))^(1/3) ≤(1/3)

$$\frac{{x}+{y}}{{z}}.\frac{{y}+{z}}{{x}}.\frac{{z}+{x}}{{y}}=\frac{\mathrm{1}−{z}}{{z}}.\frac{\mathrm{1}−{x}}{{x}}.\frac{\mathrm{1}−{y}}{{y}}= \\ $$$$\left(\frac{\mathrm{1}}{{z}}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{{y}}−\mathrm{1}\right)= \\ $$$$\left(\frac{\mathrm{1}}{{xz}}−\frac{\mathrm{1}}{{z}}−\frac{\mathrm{1}}{{x}}+\mathrm{1}\right)\left(\frac{\mathrm{1}}{{y}}−\mathrm{1}\right)= \\ $$$$=\frac{\mathrm{1}}{{xyz}}−\frac{\mathrm{1}}{{zy}}−\frac{\mathrm{1}}{{xy}}−\frac{\mathrm{1}}{{xz}}+\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}−\mathrm{1}= \\ $$$$=\frac{\mathrm{1}}{{xyz}}−\frac{{x}+{y}+{z}}{{xyz}}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}−\mathrm{1}= \\ $$$$=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}−\mathrm{1}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{xyz}}}−\mathrm{1}=\frac{\mathrm{3}}{\sqrt[{\mathrm{3}}]{{xyz}}}−\mathrm{1}\geqslant \\ $$$$\geqslant\frac{\mathrm{3}}{\frac{\mathrm{1}}{\mathrm{3}}}−\mathrm{1}=\mathrm{9}−\mathrm{1}=\mathrm{8}\:\:\:.\blacksquare \\ $$$${note}:\:\:{x}+{y}+{z}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{{xyz}}\Rightarrow\sqrt[{\mathrm{3}}]{{xyz}}\leqslant\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by mrW1 last updated on 16/May/17

I think this inequality is also valid, even if x+y+z≠1. For example x=2,y=3,z=5 ((x+y)/z)×((y+z)/x)×((z+x)/y)=(5/5)×(8/2)×(7/3)=((28)/3)>((24)/3)=8

$${I}\:{think}\:{this}\:{inequality}\:{is}\:{also}\:{valid}, \\ $$$${even}\:{if}\:{x}+{y}+{z}\neq\mathrm{1}.\:{For}\:{example}\: \\ $$$${x}=\mathrm{2},{y}=\mathrm{3},{z}=\mathrm{5} \\ $$$$\frac{{x}+{y}}{{z}}×\frac{{y}+{z}}{{x}}×\frac{{z}+{x}}{{y}}=\frac{\mathrm{5}}{\mathrm{5}}×\frac{\mathrm{8}}{\mathrm{2}}×\frac{\mathrm{7}}{\mathrm{3}}=\frac{\mathrm{28}}{\mathrm{3}}>\frac{\mathrm{24}}{\mathrm{3}}=\mathrm{8} \\ $$

Commented by Tinkutara last updated on 16/May/17

Thanks behi!

$$\mathrm{Thanks}\:\mathrm{behi}! \\ $$

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