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Question Number 108594 by Rasikh last updated on 18/Aug/20

Answered by 1549442205PVT last updated on 18/Aug/20

From a,b,c>0 (1/a)+(1/b)+(1/c)=1we have ab+bc+ca=abc(1).Apply Cauchy′s inequality for three positive numbers we have : abc=ab+bc+ca≥3^3 (√(ab.bc.ca))=3(^3 (√(abc)))^2 ⇒^3 (√(abc))≥3(1).Hence,from(1) we have P=(a−1)(b−1)(c−1)=abc−(ab+bc+ca) +(a+b+c)−1=a+b+c−1 ≥3^3 (√(abc))−1≥3.3−1=8 The equality occurs if an only if a=b=c=3 Thus,P=(a−1)(b−1)(c−1)then P_(min) =8 when (a,b,c)=(3,3,3)

$$\mathrm{From}\:\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0}\:\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{c}}=\mathrm{1we}\:\mathrm{have} \\ $$$$\mathrm{ab}+\mathrm{bc}+\mathrm{ca}=\mathrm{abc}\left(\mathrm{1}\right).\mathrm{Apply}\:\mathrm{Cauchy}'\mathrm{s}\: \\ $$$$\mathrm{inequality}\:\mathrm{for}\:\mathrm{three}\:\mathrm{positive}\:\mathrm{numbers} \\ $$$$\mathrm{we}\:\mathrm{have}\:: \\ $$$$\mathrm{abc}=\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\geqslant\mathrm{3}\:^{\mathrm{3}} \sqrt{\mathrm{ab}.\mathrm{bc}.\mathrm{ca}}=\mathrm{3}\left(\:^{\mathrm{3}} \sqrt{\mathrm{abc}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:^{\mathrm{3}} \sqrt{\mathrm{abc}}\geqslant\mathrm{3}\left(\mathrm{1}\right).\mathrm{Hence},\mathrm{from}\left(\mathrm{1}\right)\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{P}=\left(\mathrm{a}−\mathrm{1}\right)\left(\mathrm{b}−\mathrm{1}\right)\left(\mathrm{c}−\mathrm{1}\right)=\mathrm{abc}−\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right) \\ $$$$+\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)−\mathrm{1}=\mathrm{a}+\mathrm{b}+\mathrm{c}−\mathrm{1} \\ $$$$\geqslant\mathrm{3}\:^{\mathrm{3}} \sqrt{\mathrm{abc}}−\mathrm{1}\geqslant\mathrm{3}.\mathrm{3}−\mathrm{1}=\mathrm{8} \\ $$$$\mathrm{The}\:\mathrm{equality}\:\mathrm{occurs}\:\mathrm{if}\:\mathrm{an}\:\mathrm{only}\:\mathrm{if} \\ $$$$\mathrm{a}=\mathrm{b}=\mathrm{c}=\mathrm{3} \\ $$$$\mathrm{Thus},\mathrm{P}=\left(\mathrm{a}−\mathrm{1}\right)\left(\mathrm{b}−\mathrm{1}\right)\left(\mathrm{c}−\mathrm{1}\right)\mathrm{then} \\ $$$$\boldsymbol{\mathrm{P}}_{\boldsymbol{\mathrm{min}}} =\mathrm{8}\:\boldsymbol{\mathrm{when}}\:\left(\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}}\right)=\left(\mathrm{3},\mathrm{3},\mathrm{3}\right) \\ $$

Commented by Rasikh last updated on 18/Aug/20

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Answered by mathmax by abdo last updated on 18/Aug/20

we have (1/a)+(1/b)+(1/c)=1 ⇒(a/b)+(a/c) =a−1 and (b/a)+(b/c)=b−1 and (c/a)+(c/b) =c−1 ⇒(a−1)(b−1)(c−1) =((a/b) +(a/c))((b/a)+(b/c))((c/a)+(c/b)) ≥2(√(a^2 /(bc))).2(√(b^2 /(ac))).2(√(c^2 /(ab)))=8 ⇒(a−1)(b−1)(c−1)≥8 ⇒ min(...)=8

$$\mathrm{we}\:\mathrm{have}\:\:\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{c}}=\mathrm{1}\:\Rightarrow\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{a}}{\mathrm{c}}\:=\mathrm{a}−\mathrm{1}\:\mathrm{and}\:\frac{\mathrm{b}}{\mathrm{a}}+\frac{\mathrm{b}}{\mathrm{c}}=\mathrm{b}−\mathrm{1} \\ $$$$\mathrm{and}\:\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{b}}\:=\mathrm{c}−\mathrm{1}\:\Rightarrow\left(\mathrm{a}−\mathrm{1}\right)\left(\mathrm{b}−\mathrm{1}\right)\left(\mathrm{c}−\mathrm{1}\right)\:=\left(\frac{\mathrm{a}}{\mathrm{b}}\:+\frac{\mathrm{a}}{\mathrm{c}}\right)\left(\frac{\mathrm{b}}{\mathrm{a}}+\frac{\mathrm{b}}{\mathrm{c}}\right)\left(\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{b}}\right) \\ $$$$\geqslant\mathrm{2}\sqrt{\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{bc}}}.\mathrm{2}\sqrt{\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{ac}}}.\mathrm{2}\sqrt{\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{ab}}}=\mathrm{8}\:\Rightarrow\left(\mathrm{a}−\mathrm{1}\right)\left(\mathrm{b}−\mathrm{1}\right)\left(\mathrm{c}−\mathrm{1}\right)\geqslant\mathrm{8}\:\Rightarrow \\ $$$$\mathrm{min}\left(...\right)=\mathrm{8} \\ $$

Commented by Rasikh last updated on 18/Aug/20

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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