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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

$(1+\dfrac {a}{b})(1+\dfrac {b}{c})(1+\dfrac {c}{a})\geq 2(1+\dfrac {a+b+c}{\sqrt[3]{abc}})$

- Thread starter Albert
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- Thread starter
- #1

- Jan 25, 2013

- 1,225

$(1+\dfrac {a}{b})(1+\dfrac {b}{c})(1+\dfrac {c}{a})\geq 2(1+\dfrac {a+b+c}{\sqrt[3]{abc}})$

- Thread starter
- #2

- Jan 25, 2013

- 1,225

using the AM-GM inequality for left side

$(1+\dfrac {a}{b})(1+\dfrac {b}{c})(1+\dfrac {c}{a})\geq 2(1+\dfrac {a+b+c}{\sqrt[3]{abc}})$

expansion of left side =$2+k=2+\dfrac{k}{3}+\dfrac {2k}{3}\geq 4+\dfrac {2k}{3}$

where $k=\dfrac {b}{a} +\dfrac{a}{b}+\dfrac{c}{b} +\dfrac{b}{c}+\dfrac{a}{c} +\dfrac{c}{a}\geq 6 $

using the GM-HM inequality for right side

right side :$\leq 2(1+\dfrac {(a+b+c)}{\dfrac {3}{(1/a) + (1/b) + (1/c)}} =

2(1+\dfrac {(a+b+c)\times \left [(1/a) + (1/b) + (1/c)\right ]}{3} )=4+\dfrac {2k}{3}$

$\therefore $ the proof is done

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