MHB Can This Algebraic Inequality Be Proven Using AM-GM and GM-HM Methods?

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The inequality $(1+\dfrac {a}{b})(1+\dfrac {b}{c})(1+\dfrac {c}{a})\geq 2(1+\dfrac {a+b+c}{\sqrt[3]{abc}})$ is proven using the AM-GM inequality for the left side and the GM-HM inequality for the right side. The left side expands to $2 + k$ with $k$ representing the sum of ratios, which is shown to be at least 6. The right side is simplified to $2(1+\dfrac {(a+b+c)}{\dfrac {3}{(1/a) + (1/b) + (1/c)}})$, ultimately leading to a similar form. Both sides are shown to be equal, confirming the inequality holds true. Thus, the proof is successfully completed.
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a,b,c >0 , prove that :

$(1+\dfrac {a}{b})(1+\dfrac {b}{c})(1+\dfrac {c}{a})\geq 2(1+\dfrac {a+b+c}{\sqrt[3]{abc}})$
 
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Re: Prove an inequality

Albert said:
a,b,c >0 , prove that :
$(1+\dfrac {a}{b})(1+\dfrac {b}{c})(1+\dfrac {c}{a})\geq 2(1+\dfrac {a+b+c}{\sqrt[3]{abc}})$
using the AM-GM inequality for left side
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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