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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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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