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Suppose-that-a-b-c-gt-0-Prove-that-1-a-1-b-1-b-1-c-1-c-1-a-3-1-abc-

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Question Number 5380 by 314159 last updated on 12/May/16
Suppose that a,b,c>0.Prove that (1/(a(1+b)))+(1/(b(1+c)))+(1/(c(1+a))) ≥(3/(1+abc)).
Supposethata,b,c>0.Provethat1a(1+b)+1b(1+c)+1c(1+a)⩾31+abc.
Commented by Rasheed Soomro last updated on 14/May/16
LHS=((bc(1+c)(1+a)+ac(1+b)(1+a)+ab(1+b)(1+c))/(abc(1+a)(1+b)(1+c))) =((ab(1+c+b+bc)+bc(1+a+c+ca)+ca(1+a+b+ab))/(abc(1+a)(1+b)(1+c))) =((ab+abc+ab^2 +ab^2 c+bc+abc+bc^2 +abc^2 +ca+ca^2 +abc+a^2 bc)/(abc(1+a)(1+b)(1+c))) ((3abc+ab+bc+ca+ab^2 +bc^2 +ca^2 +a^2 bc+ab^2 c+abc^2 )/(abc(1+a)(1+b)(1+c))) Continue
LHS=bc(1+c)(1+a)+ac(1+b)(1+a)+ab(1+b)(1+c)abc(1+a)(1+b)(1+c)=ab(1+c+b+bc)+bc(1+a+c+ca)+ca(1+a+b+ab)abc(1+a)(1+b)(1+c)=ab+abc+ab2+ab2c+bc+abc+bc2+abc2+ca+ca2+abc+a2bcabc(1+a)(1+b)(1+c)3abc+ab+bc+ca+ab2+bc2+ca2+a2bc+ab2c+abc2abc(1+a)(1+b)(1+c)Continue
Commented by Yozzii last updated on 15/May/16
let u=(1/(a(1+b)))>0,v=(1/(b(1+c)))>0,w=(1/(c(1+a)))>0 ⇒by AM−GM ((u+v+w)/3)≥(uvw)^(1/3) ⇒u+v+w≥3(uvw)^(1/3) =3((1/(abc(1+a)(1+b)(1+c))))^(1/3) Let j=3(uvw)^(1/3) =(3/((abc(1+ab+bc+ac+a+b+c+abc))^(1/3) )) Let r=((3/(1+abc)))^3 =((27)/((1+abc)^3 ))=((27)/(1+3abc+3(abc)^2 +(abc)^3 )) j^3 =((27)/(abc+(abc)(ab+bc+ac)+(abc)(a+b+c)+(abc)^2 )) ∴(1/r)−(1/j^3 )=(((abc)^3 +3(abc)^2 +3(abc)+1−abc(1+ab+bc+ac+a+b+c+abc))/(27)) (1/r)−(1/j^3 )=(((1+abc)^3 −(a+a^2 )(b+b^2 )(c+c^2 ))/(27)) (1/r)−(1/j^3 )=(((1+abc)(1+abc)(1+abc)−abc(1+a)(1+b)(1+c))/(27)) r^(−1) −j^(−3) =((1−a^2 bc^2 +abc(1−a)+abc(1−b)+abc^2 (ab−1)+a^2 b^2 c(c−1)+ab^2 c^2 (a^2 bc−1))/(27)) (1/r)−(1/j^3 )=(((abc)^(−1) +2abc+2+(abc)^2 −a−b−c−ab−bc−ac)/(27(abc)^(−1) )) Is it true that for a,b,c>0 (1+abc)^3 ≥abc(1+a)(1+b)(1+c) ?
letu=1a(1+b)>0,v=1b(1+c)>0,w=1c(1+a)>0⇒byAM−GMu+v+w3⩾(uvw)1/3⇒u+v+w⩾3(uvw)1/3=3(1abc(1+a)(1+b)(1+c))1/3Letj=3(uvw)1/3=3(abc(1+ab+bc+ac+a+b+c+abc))1/3Letr=(31+abc)3=27(1+abc)3=271+3abc+3(abc)2+(abc)3j3=27abc+(abc)(ab+bc+ac)+(abc)(a+b+c)+(abc)2∴1r−1j3=(abc)3+3(abc)2+3(abc)+1−abc(1+ab+bc+ac+a+b+c+abc)271r−1j3=(1+abc)3−(a+a2)(b+b2)(c+c2)271r−1j3=(1+abc)(1+abc)(1+abc)−abc(1+a)(1+b)(1+c)27r−1−j−3=1−a2bc2+abc(1−a)+abc(1−b)+abc2(ab−1)+a2b2c(c−1)+ab2c2(a2bc−1)271r−1j3=(abc)−1+2abc+2+(abc)2−a−b−c−ab−bc−ac27(abc)−1Isittruethatfora,b,c>0(1+abc)3⩾abc(1+a)(1+b)(1+c)?

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