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Question Number 42492 by maxmathsup by imad last updated on 26/Aug/18

let x>0 ,y>0,z>0   prove that  (x^2 /(yz)) +(y^2 /(xz)) +(z^2 /(xy)) ≥3 .

$${let}\:{x}>\mathrm{0}\:,{y}>\mathrm{0},{z}>\mathrm{0}\:\:\:{prove}\:{that}\:\:\frac{{x}^{\mathrm{2}} }{{yz}}\:+\frac{{y}^{\mathrm{2}} }{{xz}}\:+\frac{{z}^{\mathrm{2}} }{{xy}}\:\geqslant\mathrm{3}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Aug/18

(x^3 /(xyz))+(y^3 /(xyz))+(z^3 /(xyz))−3  ((x^3 +y^3 +z^3 −3xyz)/(xyz))  (((x+y+z)(x^2 +y^2 +z^2 −xy−yz−zx))/(xyz))  (((x+y+z){(x−y)^2 +(y−z)^2 +(z−x)^2 })/(2xyz))  +ve   (since x>0   y>0    z>0)  so  (x^2 /(yz))+(y^2 /(xz))+(z^2 /(xy))>3 proved  now if (x=y=z  then (x^2 /(yz))+(y^2 /(xz))+(z^2 /(xy))−3=0  hence (x^2 /(yz))+(y^2 /(zx))+(z^2 /(xy))≥3  proved

$$\frac{{x}^{\mathrm{3}} }{{xyz}}+\frac{{y}^{\mathrm{3}} }{{xyz}}+\frac{{z}^{\mathrm{3}} }{{xyz}}−\mathrm{3} \\ $$ $$\frac{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} −\mathrm{3}{xyz}}{{xyz}} \\ $$ $$\frac{\left({x}+{y}+{z}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −{xy}−{yz}−{zx}\right)}{{xyz}} \\ $$ $$\frac{\left({x}+{y}+{z}\right)\left\{\left({x}−{y}\right)^{\mathrm{2}} +\left({y}−{z}\right)^{\mathrm{2}} +\left({z}−{x}\right)^{\mathrm{2}} \right\}}{\mathrm{2}{xyz}} \\ $$ $$+{ve}\:\:\:\left({since}\:{x}>\mathrm{0}\:\:\:{y}>\mathrm{0}\:\:\:\:{z}>\mathrm{0}\right) \\ $$ $${so}\:\:\frac{{x}^{\mathrm{2}} }{{yz}}+\frac{{y}^{\mathrm{2}} }{{xz}}+\frac{{z}^{\mathrm{2}} }{{xy}}>\mathrm{3}\:{proved} \\ $$ $${now}\:{if}\:\left({x}={y}={z}\right. \\ $$ $${then}\:\frac{{x}^{\mathrm{2}} }{{yz}}+\frac{{y}^{\mathrm{2}} }{{xz}}+\frac{{z}^{\mathrm{2}} }{{xy}}−\mathrm{3}=\mathrm{0} \\ $$ $${hence}\:\frac{{x}^{\mathrm{2}} }{{yz}}+\frac{{y}^{\mathrm{2}} }{{zx}}+\frac{{z}^{\mathrm{2}} }{{xy}}\geqslant\mathrm{3}\:\:{proved} \\ $$

Commented bymath khazana by abdo last updated on 27/Aug/18

correct sir tanmay thanks.

$${correct}\:{sir}\:{tanmay}\:{thanks}. \\ $$

Commented bytanmay.chaudhury50@gmail.com last updated on 27/Aug/18

its ok sir

$${its}\:{ok}\:{sir} \\ $$

Answered by behi83417@gmail.com last updated on 26/Aug/18

(x^2 /(yz))+(y^2 /(xz))+(z^2 /(xy))≥3(((x^2 /(yz)).(y^2 /(xz)).(z^2 /(xy))))^(1/3) =3(1)^(1/3) =3.

$$\frac{{x}^{\mathrm{2}} }{{yz}}+\frac{{y}^{\mathrm{2}} }{{xz}}+\frac{{z}^{\mathrm{2}} }{{xy}}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\frac{{x}^{\mathrm{2}} }{{yz}}.\frac{{y}^{\mathrm{2}} }{{xz}}.\frac{{z}^{\mathrm{2}} }{{xy}}}=\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{1}}=\mathrm{3}. \\ $$

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