Question Number 109178 by Algoritm last updated on 21/Aug/20
Commented by Algoritm last updated on 21/Aug/20
$$\mathrm{mr}\:\mathrm{W} \\ $$
Answered by Dwaipayan Shikari last updated on 21/Aug/20
Commented by Algoritm last updated on 21/Aug/20
$$??? \\ $$
Commented by aurpeyz last updated on 21/Aug/20
$${wow}.\:{How}\:{do}\:{you}\:{get}\:{to}\:{think}\:{of}\:{this}\:{process}?\: \\ $$$${i}\:{like}\:{it} \\ $$
Commented by floor(10²Eta[1]) last updated on 22/Aug/20
$$\mathrm{he}\:\mathrm{just}\:\mathrm{used}\:\mathrm{the}\:\mathrm{AM}-\mathrm{HM}\:\mathrm{inequality}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{AM}-\mathrm{GM}\:\mathrm{inequality} \\ $$
Answered by 1549442205PVT last updated on 22/Aug/20
![We have: (x+y+z)((1/x)+(1/y)+(1/z))+((8xyz)/((x+y)(y+z)(z+x)))≥10 ⇔(((x+y+z)(xy+yz+zx))/(xyz))+((8xyz)/((x+y)(y+z)(z+x)))≥10(1) First,we note some following identities: (x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)−xyz Hence,The inequality (1) is equivalent to (((x+y+z)(xy+yz+zx))/(xyz))+((8xyz)/((x+y+z)(xy+yz+zx)−xyz))≥10(2) Since the nominator and denomirator of each the fraction in the LHS are in same degree (degree 3),WLOG we can suppose that x+y+z=3(because if x+y+z=m then put x=((am)/3),y=((bm)/3), z=((cm)/3)⇒((am)/3)+((bm)/3)+((cm)/3)=m⇒a+b+c=3 and the inequality (2) becomes (((m/3)^3 (a+b+c)(ab+bc+ca))/((m/3)^3 abc))+((8(m/3)^3 abc)/((m/3)^3 [(a+b+c)(ab+bc+ca)−abc]))≥10 ⇔(( (a+b+c)(ab+bc+ca))/(abc))+((8abc)/((a+b+c)(ab+bc+ca)−abc))≥10 which is also the inequality (2) but under the condition a+b+c=3) By that reason we have (2)⇔((3(xy+yz+zx))/(xyz))+((8xyz)/(3(xy+yz+zx)−xyz))≥10 ⇔9(xy+yz+zx)^2 −3xyz(xy+yz+zx) +18(xyz)^2 ≥10xyz[3(xy+yz+zx)−xyz] ⇔9(xy+yz+zx)^2 +18(xyz)^2 ≥ 33xyz(xy+yz+zx)(3) Now we put xy=m,xz=n,yz=p.Then 3^2 =(x+y+z)^2 ≥3(xy+yz+zx) ⇒m+n+p=xy+yz+zx≤3 3≥xy+yz+zx≥3^3 (√(xy.yz.zx))=3^3 (√((xyz)^2 )) ⇒^3 (√(xyz)) ≤1,mnp≤(((m+m+p)/3))^3 ≤1(∗) We have also (xyz)^2 =mnp.Therefore, (3)⇔9(m+n+p)^2 +18mnp≥33(m+n+p)(√(mnp)) (3′) Apply Cauchy′s inequality for two positive numbers we have 2(m+n+p)^2 +18mnp≥(√(2(m+n+p)^2 .8mnp)) =8(m+n+p)(√(mnp)) (i).On the other hands, Since mnp≤1⇒(√(mnp)) ≤1(due to (∗)) ⇒mnp(√(mnp))=(√((mnp)^3 ))≤mnp ⇒(√(mnp))≤^3 (√(mnp)) (∗∗).Now apply Cauchy′s inequality for three positive numbers we get: m^2 +mn+mp≥3^3 (√(m^2 .mn.mp)) =3m^3 (√(mnp)) ≥3(√(mnp)) (due to (∗∗)).Similarly, n^2 +nm+np≥3n(√(mnp)) p^2 +pn+pm≥3p(√(mnp)) Adding up three above inequalities we get m^2 +n^2 +p^2 +2(mn+mp+np) =(m+n+p)^2 ≥3(m+n+p)(√(mnp)) ⇒7(m+n+p)^3 ≥21(m+n+p)(√(mnp))(ii) Adding up two inequalities (i)and(ii) we get: 9(m+n+p)^2 +18mnp≥33(m+n+p)(√(mnp)) ,so the inequality (3′)is true which implies that (2) is true .Consequently, the inequality (1)is proved.Thus,we proved the inequality: (x+y+z)((1/x)+(1/y)+(1/z))+((8xyz)/((x+y)(y+z)(z+x)))≥10 The equality ocurrs if and only if x=y=z (Q.E.D)](https://www.tinkutara.com/question/Q109252.png)
$$\mathrm{We}\:\mathrm{have}: \\ $$$$\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left(\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{y}}+\frac{\mathrm{1}}{\mathrm{z}}\right)+\frac{\mathrm{8xyz}}{\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{y}+\mathrm{z}\right)\left(\mathrm{z}+\mathrm{x}\right)}\geqslant\mathrm{10} \\ $$$$\Leftrightarrow\frac{\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)}{\mathrm{xyz}}+\frac{\mathrm{8xyz}}{\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{y}+\mathrm{z}\right)\left(\mathrm{z}+\mathrm{x}\right)}\geqslant\mathrm{10}\left(\mathrm{1}\right) \\ $$$$\mathrm{First},\mathrm{we}\:\mathrm{note}\:\mathrm{some}\:\mathrm{following}\:\mathrm{identities}: \\ $$$$\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{y}+\mathrm{z}\right)\left(\mathrm{z}+\mathrm{x}\right)=\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)−\mathrm{xyz} \\ $$$$\mathrm{Hence},\mathrm{The}\:\mathrm{inequality}\:\left(\mathrm{1}\right)\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to} \\ $$$$\frac{\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)}{\mathrm{xyz}}+\frac{\mathrm{8xyz}}{\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)−\mathrm{xyz}}\geqslant\mathrm{10}\left(\mathrm{2}\right) \\ $$$$\mathrm{Since}\:\mathrm{the}\:\mathrm{nominator}\:\mathrm{and}\:\mathrm{denomirator} \\ $$$$\mathrm{of}\:\mathrm{each}\:\mathrm{the}\:\mathrm{fraction}\:\mathrm{in}\:\mathrm{the}\:\mathrm{LHS}\:\mathrm{are} \\ $$$$\mathrm{in}\:\mathrm{same}\:\mathrm{degree}\:\left(\mathrm{degree}\:\mathrm{3}\right),\mathrm{WLOG}\:\mathrm{we} \\ $$$$\mathrm{can}\:\mathrm{suppose}\:\mathrm{that}\:\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{3}\left(\mathrm{because}\right. \\ $$$$\mathrm{if}\:\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{m}\:\mathrm{then}\:\mathrm{put}\:\mathrm{x}=\frac{\mathrm{am}}{\mathrm{3}},\mathrm{y}=\frac{\mathrm{bm}}{\mathrm{3}}, \\ $$$$\mathrm{z}=\frac{\mathrm{cm}}{\mathrm{3}}\Rightarrow\frac{\mathrm{am}}{\mathrm{3}}+\frac{\mathrm{bm}}{\mathrm{3}}+\frac{\mathrm{cm}}{\mathrm{3}}=\mathrm{m}\Rightarrow\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{3} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{inequality}\:\left(\mathrm{2}\right)\:\mathrm{becomes} \\ $$$$\frac{\left(\mathrm{m}/\mathrm{3}\right)^{\mathrm{3}} \left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)}{\left(\mathrm{m}/\mathrm{3}\right)^{\mathrm{3}} \mathrm{abc}}+\frac{\mathrm{8}\left(\mathrm{m}/\mathrm{3}\right)^{\mathrm{3}} \mathrm{abc}}{\left(\mathrm{m}/\mathrm{3}\right)^{\mathrm{3}} \left[\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)−\mathrm{abc}\right]}\geqslant\mathrm{10} \\ $$$$\Leftrightarrow\frac{\:\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)}{\mathrm{abc}}+\frac{\mathrm{8abc}}{\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)−\mathrm{abc}}\geqslant\mathrm{10} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{also}\:\mathrm{the}\:\mathrm{inequality}\:\left(\mathrm{2}\right)\:\mathrm{but}\: \\ $$$$\left.\mathrm{under}\:\mathrm{the}\:\mathrm{condition}\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{3}\right) \\ $$$$\mathrm{By}\:\mathrm{that}\:\mathrm{reason}\:\mathrm{we}\:\mathrm{have} \\ $$$$\left(\mathrm{2}\right)\Leftrightarrow\frac{\mathrm{3}\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)}{\mathrm{xyz}}+\frac{\mathrm{8xyz}}{\mathrm{3}\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)−\mathrm{xyz}}\geqslant\mathrm{10} \\ $$$$\Leftrightarrow\mathrm{9}\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)^{\mathrm{2}} −\mathrm{3xyz}\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right) \\ $$$$+\mathrm{18}\left(\mathrm{xyz}\right)^{\mathrm{2}} \geqslant\mathrm{10xyz}\left[\mathrm{3}\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)−\mathrm{xyz}\right] \\ $$$$\Leftrightarrow\mathrm{9}\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)^{\mathrm{2}} +\mathrm{18}\left(\mathrm{xyz}\right)^{\mathrm{2}} \geqslant \\ $$$$\mathrm{33xyz}\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)\left(\mathrm{3}\right) \\ $$$$\mathrm{Now}\:\mathrm{we}\:\mathrm{put}\:\mathrm{xy}=\mathrm{m},\mathrm{xz}=\mathrm{n},\mathrm{yz}=\mathrm{p}.\mathrm{Then} \\ $$$$\mathrm{3}^{\mathrm{2}} =\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)^{\mathrm{2}} \geqslant\mathrm{3}\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right) \\ $$$$\Rightarrow\mathrm{m}+\mathrm{n}+\mathrm{p}=\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\leqslant\mathrm{3} \\ $$$$\mathrm{3}\geqslant\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\geqslant\mathrm{3}\:^{\mathrm{3}} \sqrt{\mathrm{xy}.\mathrm{yz}.\mathrm{zx}}=\mathrm{3}\:^{\mathrm{3}} \sqrt{\left(\mathrm{xyz}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:^{\mathrm{3}} \sqrt{\mathrm{xyz}}\:\leqslant\mathrm{1},\mathrm{mnp}\leqslant\left(\frac{\mathrm{m}+\mathrm{m}+\mathrm{p}}{\mathrm{3}}\right)^{\mathrm{3}} \leqslant\mathrm{1}\left(\ast\right) \\ $$$$\mathrm{We}\:\mathrm{have}\:\mathrm{also}\:\left(\mathrm{xyz}\right)^{\mathrm{2}} =\mathrm{mnp}.\mathrm{Therefore}, \\ $$$$\left(\mathrm{3}\right)\Leftrightarrow\mathrm{9}\left(\mathrm{m}+\mathrm{n}+\mathrm{p}\right)^{\mathrm{2}} +\mathrm{18mnp}\geqslant\mathrm{33}\left(\mathrm{m}+\mathrm{n}+\mathrm{p}\right)\sqrt{\mathrm{mnp}}\:\left(\mathrm{3}'\right) \\ $$$$\mathrm{Apply}\:\:\mathrm{Cauchy}'\mathrm{s}\:\mathrm{inequality}\:\mathrm{for}\:\mathrm{two} \\ $$$$\mathrm{positive}\:\mathrm{numbers}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{2}\left(\mathrm{m}+\mathrm{n}+\mathrm{p}\right)^{\mathrm{2}} +\mathrm{18mnp}\geqslant\sqrt{\mathrm{2}\left(\mathrm{m}+\mathrm{n}+\mathrm{p}\right)^{\mathrm{2}} .\mathrm{8mnp}} \\ $$$$=\mathrm{8}\left(\mathrm{m}+\mathrm{n}+\mathrm{p}\right)\sqrt{\mathrm{mnp}}\:\left(\mathrm{i}\right).\mathrm{On}\:\mathrm{the}\:\mathrm{other}\:\mathrm{hands}, \\ $$$$\mathrm{Since}\:\mathrm{mnp}\leqslant\mathrm{1}\Rightarrow\sqrt{\mathrm{mnp}}\:\leqslant\mathrm{1}\left(\mathrm{due}\:\mathrm{to}\:\left(\ast\right)\right) \\ $$$$\Rightarrow\mathrm{mnp}\sqrt{\mathrm{mnp}}=\sqrt{\left(\mathrm{mnp}\right)^{\mathrm{3}} }\leqslant\mathrm{mnp} \\ $$$$\Rightarrow\sqrt{\mathrm{mnp}}\leqslant\:^{\mathrm{3}} \sqrt{\mathrm{mnp}}\:\:\:\:\left(\ast\ast\right).\mathrm{Now}\:\mathrm{apply} \\ $$$$\mathrm{Cauchy}'\mathrm{s}\:\mathrm{inequality}\:\mathrm{for}\:\mathrm{three}\:\mathrm{positive} \\ $$$$\mathrm{numbers}\:\mathrm{we}\:\mathrm{get}: \\ $$$$\mathrm{m}^{\mathrm{2}} +\mathrm{mn}+\mathrm{mp}\geqslant\mathrm{3}\:^{\mathrm{3}} \sqrt{\mathrm{m}^{\mathrm{2}} .\mathrm{mn}.\mathrm{mp}}\:=\mathrm{3m}\:^{\mathrm{3}} \sqrt{\mathrm{mnp}} \\ $$$$\geqslant\mathrm{3}\sqrt{\mathrm{mnp}}\:\:\:\left(\mathrm{due}\:\mathrm{to}\:\left(\ast\ast\right)\right).\mathrm{Similarly}, \\ $$$$\mathrm{n}^{\mathrm{2}} +\mathrm{nm}+\mathrm{np}\geqslant\mathrm{3n}\sqrt{\mathrm{mnp}} \\ $$$$\mathrm{p}^{\mathrm{2}} +\mathrm{pn}+\mathrm{pm}\geqslant\mathrm{3p}\sqrt{\mathrm{mnp}} \\ $$$$\mathrm{Adding}\:\mathrm{up}\:\mathrm{three}\:\mathrm{above}\:\mathrm{inequalities} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} +\mathrm{p}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{mn}+\mathrm{mp}+\mathrm{np}\right) \\ $$$$=\left(\mathrm{m}+\mathrm{n}+\mathrm{p}\right)^{\mathrm{2}} \geqslant\mathrm{3}\left(\mathrm{m}+\mathrm{n}+\mathrm{p}\right)\sqrt{\mathrm{mnp}} \\ $$$$\Rightarrow\mathrm{7}\left(\mathrm{m}+\mathrm{n}+\mathrm{p}\right)^{\mathrm{3}} \geqslant\mathrm{21}\left(\mathrm{m}+\mathrm{n}+\mathrm{p}\right)\sqrt{\mathrm{mnp}}\left(\mathrm{ii}\right) \\ $$$$\mathrm{Adding}\:\mathrm{up}\:\mathrm{two}\:\mathrm{inequalities}\:\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right) \\ $$$$\mathrm{we}\:\mathrm{get}: \\ $$$$\mathrm{9}\left(\mathrm{m}+\mathrm{n}+\mathrm{p}\right)^{\mathrm{2}} +\mathrm{18mnp}\geqslant\mathrm{33}\left(\mathrm{m}+\mathrm{n}+\mathrm{p}\right)\sqrt{\mathrm{mnp}} \\ $$$$,\mathrm{so}\:\mathrm{the}\:\mathrm{inequality}\:\left(\mathrm{3}'\right)\mathrm{is}\:\mathrm{true}\:\mathrm{which} \\ $$$$\mathrm{implies}\:\mathrm{that}\:\left(\mathrm{2}\right)\:\mathrm{is}\:\mathrm{true}\:.\mathrm{Consequently}, \\ $$$$\mathrm{the}\:\mathrm{inequality}\:\left(\mathrm{1}\right)\mathrm{is}\:\mathrm{proved}.\mathrm{Thus},\mathrm{we} \\ $$$$\mathrm{proved}\:\mathrm{the}\:\mathrm{inequality}: \\ $$$$\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{z}}\right)\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{z}}}\right)+\frac{\mathrm{8}\boldsymbol{\mathrm{xyz}}}{\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\right)\left(\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{z}}\right)\left(\boldsymbol{\mathrm{z}}+\boldsymbol{\mathrm{x}}\right)}\geqslant\mathrm{10} \\ $$$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{equality}}\:\boldsymbol{\mathrm{ocurrs}}\:\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{only}}\:\boldsymbol{\mathrm{if}} \\ $$$$\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{z}}\:\left(\boldsymbol{\mathrm{Q}}.\boldsymbol{\mathrm{E}}.\boldsymbol{\mathrm{D}}\right)\: \\ $$