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problem: a,b,c>0 and $abc=1$ prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a+b+c$

my attempt:

$LHS=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=(a+b+c) \frac{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}{a+b+c}$

From weighted AM-GM;

$LHS\geq (a+b+c) ((\frac{1}{b})^a(\frac{1}{c})^b(\frac{1}{a})^c)^{\frac{1}{a+b+c}}$

and we have :

$ ((\frac{1}{b})^a(\frac{1}{c})^b(\frac{1}{a})^c)^{\frac{1}{a+b+c}}\geq 1$

because: if $a\leq b \leq c $ :then $((\frac{1}{b})^a(\frac{1}{c})^b(\frac{1}{a})^c)\geq ((\frac{1}{b})^a(\frac{1}{c})^a(\frac{1}{a})^a)=(\frac{1}{abc})^a=1$

and if $c\leq a \leq b $ :then $((\frac{1}{b})^a(\frac{1}{c})^b(\frac{1}{a})^c)\geq ((\frac{1}{b})^c(\frac{1}{c})^c(\frac{1}{a})^c)=(\frac{1}{abc})^c=1$.and the same for others case

so finally:

$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=(a+b+c) \frac{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}{a+b+c}\geq a+b+c.$

question:

-does my attempt is true?

  • are you sur????, i did my attempt and i asked for verifie if it wrong or true?IF YOU LOOK AT THE QUESTION YOU WILL SEE "does my attempt is true?" not "help "!!!!!, so any question else it is different than my question !!!! –  Jul 03 '22 at 17:14
  • why it closed!!!!!! please read the qusetion before closing –  Jul 03 '22 at 17:17
  • iam waiting for your explanation about this "close"!! –  Jul 03 '22 at 17:19
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    Maybe your question is closed because this solution is also just a various solution to the question above. The inequality "Weighted AM-GM" is just an expansion of AM-GM inequality, so you can just modify it to AM-GM inequality. So, this approach is similar to RijulSaini and FrancescoSica's answer to the duplicate question. (In fact, we can say that their answer is more similar.) – RDK Jul 04 '22 at 03:24
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    I recommend you to put this as an answer to the duplicate question. – RDK Jul 04 '22 at 03:48
  • @RDK thank you, –  Jul 04 '22 at 06:00

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