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mathslover

  • 3 years ago

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

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  1. mathslover
    • 3 years ago
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    @hartnn

  2. mathslover
    • 3 years ago
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    @Zarkon and @myininaya

  3. hartnn
    • 3 years ago
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    sure its difficult ?

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

  5. anonymous
    • 3 years ago
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    i suck at these

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

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

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

  9. anonymous
    • 3 years ago
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    hello joe!!!

  10. joemath314159
    • 3 years ago
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    hey :)

  11. anonymous
    • 3 years ago
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    i was thinking harmonic mean and geometric mean or something but i bet i was wrong

  12. anonymous
    • 3 years ago
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    long time, how you been?

  13. joemath314159
    • 3 years ago
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    not bad, its senior year, so work work work lol

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

  15. anonymous
    • 3 years ago
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    keep busy, glad to see you!

  16. joemath314159
    • 3 years ago
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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.

  17. joemath314159
    • 3 years ago
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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.

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

  19. mathslover
    • 3 years ago
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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) \[\large{\frac{a+b+c}{3} \times \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3} \ge (abc)^{\frac{1}{3}} \times (\frac{1}{abc})^{\frac{1}{3}}}\] \[\large{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge 9}\]

  20. mathslover
    • 3 years ago
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    And.,.... Hence proved.. Right @joemath314159 ?

  21. joemath314159
    • 3 years ago
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    very interesting :)

  22. mathslover
    • 3 years ago
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    :) yes .

  23. hartnn
    • 3 years ago
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    finally u are doing only AM>=HM only

  24. hartnn
    • 3 years ago
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    but other way round......

  25. mathslover
    • 3 years ago
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    yep :)

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