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

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SUMMARY

The algebraic 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 and GM-HM inequality. The left side is expanded to $2+k=2+\dfrac{k}{3}+\dfrac {2k}{3}$, where $k=\dfrac {b}{a} +\dfrac{a}{b}+\dfrac{c}{b} +\dfrac{b}{c}+\dfrac{a}{c} +\dfrac{c}{a}\geq 6$. The right side is shown to be less than or equal to $2(1+\dfrac {(a+b+c)\times \left [(1/a) + (1/b) + (1/c)\right ]}{3})=4+\dfrac {2k}{3}$. Thus, the proof is complete.

PREREQUISITES
  • Understanding of AM-GM (Arithmetic Mean-Geometric Mean) inequality
  • Familiarity with GM-HM (Geometric Mean-Harmonic Mean) inequality
  • Basic algebraic manipulation skills
  • Knowledge of inequalities involving positive real numbers
NEXT STEPS
  • Study the applications of AM-GM inequality in various mathematical proofs
  • Explore advanced topics in inequalities, such as Cauchy-Schwarz inequality
  • Learn about the properties and applications of GM-HM inequality
  • Investigate other methods for proving inequalities in algebra
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Mathematicians, students studying inequalities, and anyone interested in advanced algebraic proofs will benefit from this discussion.

Albert1
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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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