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mathslover Group Title

if : a > 0 , b > 0 and c>0 , prove that: \(\large{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge 9}\)

  • 2 years ago
  • 2 years ago

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  1. mathslover Group Title
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    @hartnn

    • 2 years ago
  2. mathslover Group Title
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    @Zarkon and @myininaya

    • 2 years ago
  3. hartnn Group Title
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    sure its difficult ?

    • 2 years ago
  4. mathslover Group Title
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    Unless you know the first step it is difficult. :)

    • 2 years ago
  5. satellite73 Group Title
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    i suck at these

    • 2 years ago
  6. mathslover Group Title
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    No worries, either the viewers will let you out from this question or I will at the end.

    • 2 years ago
  7. mathslover Group Title
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    Here are many legends viewing this, I would like a better answer from them.

    • 2 years ago
  8. joemath314159 Group Title
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    if x >0, then:\[x+\frac{1}{x}\ge2\]you can use this to show the desired inequality.

    • 2 years ago
  9. satellite73 Group Title
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    hello joe!!!

    • 2 years ago
  10. joemath314159 Group Title
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    hey :)

    • 2 years ago
  11. satellite73 Group Title
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    i was thinking harmonic mean and geometric mean or something but i bet i was wrong

    • 2 years ago
  12. satellite73 Group Title
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    long time, how you been?

    • 2 years ago
  13. joemath314159 Group Title
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    not bad, its senior year, so work work work lol

    • 2 years ago
  14. mathslover Group Title
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    @joemath314159 how? I am curious to know that how will it work?

    • 2 years ago
  15. satellite73 Group Title
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    keep busy, glad to see you!

    • 2 years ago
  16. joemath314159 Group Title
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    So \[(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=3+(\frac{a}{b}+\frac{b}{a})+(\frac{a}{c}+\frac{c}{a})+(\frac{c}{b}+\frac{b}{c})\]Since a,b,c are all positive, the fractions are all positive. That inequality i posted applies to all three parenthesis.

    • 2 years ago
  17. joemath314159 Group Title
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    if\[x=\frac{a}{b}\]then\[\frac{a}{b}+\frac{b}{a}=x+\frac{1}{x}\]which is greater than 2. Similarly for the other guys.

    • 2 years ago
  18. hartnn Group Title
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    AM>=HM (a+b+c)/3 >= 3/(1/a+1/b+1/c)

    • 2 years ago
  19. mathslover Group Title
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    hmn you have a right solution there but.... I did like this: The three numbers are like this... \(a,b,c\) \(AM \ge GM\) \(\large{\frac{(a+b+c)}{3} \ge (abc)^{\frac{1}{3}}}\) ---1) If three numbers are .. \(\large{\frac{1}{a}. \frac{1}{b}, \frac{1}{c}}\) \(AM\ge GM\) \(\large{\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3} \ge (\frac{1}{abc})^{(\frac{1}{3})}}\)---2) Multiply 1) and 2) \[\large{\frac{a+b+c}{3} \times \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3} \ge (abc)^{\frac{1}{3}} \times (\frac{1}{abc})^{\frac{1}{3}}}\] \[\large{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge 9}\]

    • 2 years ago
  20. mathslover Group Title
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    And.,.... Hence proved.. Right @joemath314159 ?

    • 2 years ago
  21. joemath314159 Group Title
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    very interesting :)

    • 2 years ago
  22. mathslover Group Title
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    :) yes .

    • 2 years ago
  23. hartnn Group Title
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    finally u are doing only AM>=HM only

    • 2 years ago
  24. hartnn Group Title
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    but other way round......

    • 2 years ago
  25. mathslover Group Title
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    yep :)

    • 2 years ago
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