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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)).
$${Suppose}\:{that}\:{a},{b},{c}>\mathrm{0}.{Prove}\:{that}\: \\ $$$$\frac{\mathrm{1}}{{a}\left(\mathrm{1}+{b}\right)}+\frac{\mathrm{1}}{{b}\left(\mathrm{1}+{c}\right)}+\frac{\mathrm{1}}{{c}\left(\mathrm{1}+{a}\right)}\:\geqslant\frac{\mathrm{3}}{\mathrm{1}+{abc}}. \\ $$
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
$$\mathrm{LHS}=\frac{{bc}\left(\mathrm{1}+{c}\right)\left(\mathrm{1}+{a}\right)+{ac}\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{a}\right)+{ab}\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)}{{a}\mathrm{bc}\left(\mathrm{1}+\mathrm{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+\mathrm{c}\right)} \\ $$$$=\frac{{ab}\left(\mathrm{1}+{c}+{b}+{bc}\right)+{bc}\left(\mathrm{1}+{a}+{c}+{ca}\right)+{ca}\left(\mathrm{1}+{a}+{b}+{ab}\right)}{{a}\mathrm{bc}\left(\mathrm{1}+\mathrm{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+\mathrm{c}\right)} \\ $$$$=\frac{{ab}+{abc}+{ab}^{\mathrm{2}} +{ab}^{\mathrm{2}} {c}+{bc}+{abc}+{bc}^{\mathrm{2}} +{abc}^{\mathrm{2}} +{ca}+{ca}^{\mathrm{2}} +{abc}+{a}^{\mathrm{2}} {bc}}{{a}\mathrm{bc}\left(\mathrm{1}+\mathrm{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+\mathrm{c}\right)} \\ $$$$\frac{\mathrm{3}{abc}+{ab}+{bc}+{ca}+{ab}^{\mathrm{2}} +{bc}^{\mathrm{2}} +{ca}^{\mathrm{2}} +{a}^{\mathrm{2}} {bc}+{ab}^{\mathrm{2}} {c}+{abc}^{\mathrm{2}} }{{a}\mathrm{bc}\left(\mathrm{1}+\mathrm{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+\mathrm{c}\right)} \\ $$$$\mathrm{Continue} \\ $$
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) ?
$${let}\:{u}=\frac{\mathrm{1}}{{a}\left(\mathrm{1}+{b}\right)}>\mathrm{0},{v}=\frac{\mathrm{1}}{{b}\left(\mathrm{1}+{c}\right)}>\mathrm{0},{w}=\frac{\mathrm{1}}{{c}\left(\mathrm{1}+{a}\right)}>\mathrm{0} \\ $$$$\Rightarrow{by}\:{AM}−{GM} \\ $$$$\frac{{u}+{v}+{w}}{\mathrm{3}}\geqslant\left({uvw}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\Rightarrow{u}+{v}+{w}\geqslant\mathrm{3}\left({uvw}\right)^{\mathrm{1}/\mathrm{3}} =\mathrm{3}\left(\frac{\mathrm{1}}{{abc}\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${Let}\:{j}=\mathrm{3}\left({uvw}\right)^{\mathrm{1}/\mathrm{3}} =\frac{\mathrm{3}}{\left({abc}\left(\mathrm{1}+{ab}+{bc}+{ac}+{a}+{b}+{c}+{abc}\right)\right)^{\mathrm{1}/\mathrm{3}} } \\ $$$${Let}\:{r}=\left(\frac{\mathrm{3}}{\mathrm{1}+{abc}}\right)^{\mathrm{3}} =\frac{\mathrm{27}}{\left(\mathrm{1}+{abc}\right)^{\mathrm{3}} }=\frac{\mathrm{27}}{\mathrm{1}+\mathrm{3}{abc}+\mathrm{3}\left({abc}\right)^{\mathrm{2}} +\left({abc}\right)^{\mathrm{3}} } \\ $$$${j}^{\mathrm{3}} =\frac{\mathrm{27}}{{abc}+\left({abc}\right)\left({ab}+{bc}+{ac}\right)+\left({abc}\right)\left({a}+{b}+{c}\right)+\left({abc}\right)^{\mathrm{2}} } \\ $$$$\therefore\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{{j}^{\mathrm{3}} }=\frac{\left({abc}\right)^{\mathrm{3}} +\mathrm{3}\left({abc}\right)^{\mathrm{2}} +\mathrm{3}\left({abc}\right)+\mathrm{1}−{abc}\left(\mathrm{1}+{ab}+{bc}+{ac}+{a}+{b}+{c}+{abc}\right)}{\mathrm{27}} \\ $$$$\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{{j}^{\mathrm{3}} }=\frac{\left(\mathrm{1}+{abc}\right)^{\mathrm{3}} −\left({a}+{a}^{\mathrm{2}} \right)\left({b}+{b}^{\mathrm{2}} \right)\left({c}+{c}^{\mathrm{2}} \right)}{\mathrm{27}} \\ $$$$\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{{j}^{\mathrm{3}} }=\frac{\left(\mathrm{1}+{abc}\right)\left(\mathrm{1}+{abc}\right)\left(\mathrm{1}+{abc}\right)−{abc}\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)}{\mathrm{27}} \\ $$$${r}^{−\mathrm{1}} −{j}^{−\mathrm{3}} =\frac{\mathrm{1}−{a}^{\mathrm{2}} {bc}^{\mathrm{2}} +{abc}\left(\mathrm{1}−{a}\right)+{abc}\left(\mathrm{1}−{b}\right)+{abc}^{\mathrm{2}} \left({ab}−\mathrm{1}\right)+{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}\left({c}−\mathrm{1}\right)+{ab}^{\mathrm{2}} {c}^{\mathrm{2}} \left({a}^{\mathrm{2}} {bc}−\mathrm{1}\right)}{\mathrm{27}} \\ $$$$\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{{j}^{\mathrm{3}} }=\frac{\left({abc}\right)^{−\mathrm{1}} +\mathrm{2}{abc}+\mathrm{2}+\left({abc}\right)^{\mathrm{2}} −{a}−{b}−{c}−{ab}−{bc}−{ac}}{\mathrm{27}\left({abc}\right)^{−\mathrm{1}} } \\ $$$$ \\ $$$${Is}\:{it}\:{true}\:{that}\:{for}\:{a},{b},{c}>\mathrm{0} \\ $$$$\left(\mathrm{1}+{abc}\right)^{\mathrm{3}} \geqslant{abc}\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)\:? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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