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

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mathsloverBest ResponseYou've already chosen the best response.2
@Zarkon and @myininaya
 one year ago

mathsloverBest ResponseYou've already chosen the best response.2
Unless you know the first step it is difficult. :)
 one year ago

mathsloverBest ResponseYou've already chosen the best response.2
No worries, either the viewers will let you out from this question or I will at the end.
 one year ago

mathsloverBest ResponseYou've already chosen the best response.2
Here are many legends viewing this, I would like a better answer from them.
 one year ago

joemath314159Best ResponseYou've already chosen the best response.2
if x >0, then:\[x+\frac{1}{x}\ge2\]you can use this to show the desired inequality.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
i was thinking harmonic mean and geometric mean or something but i bet i was wrong
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
long time, how you been?
 one year ago

joemath314159Best ResponseYou've already chosen the best response.2
not bad, its senior year, so work work work lol
 one year ago

mathsloverBest ResponseYou've already chosen the best response.2
@joemath314159 how? I am curious to know that how will it work?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
keep busy, glad to see you!
 one year ago

joemath314159Best ResponseYou've already chosen the best response.2
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.
 one year ago

joemath314159Best ResponseYou've already chosen the best response.2
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.
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
AM>=HM (a+b+c)/3 >= 3/(1/a+1/b+1/c)
 one year ago

mathsloverBest ResponseYou've already chosen the best response.2
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) \[\large{\frac{a+b+c}{3} \times \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3} \ge (abc)^{\frac{1}{3}} \times (\frac{1}{abc})^{\frac{1}{3}}}\] \[\large{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge 9}\]
 one year ago

mathsloverBest ResponseYou've already chosen the best response.2
And.,.... Hence proved.. Right @joemath314159 ?
 one year ago

joemath314159Best ResponseYou've already chosen the best response.2
very interesting :)
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
finally u are doing only AM>=HM only
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
but other way round......
 one year ago
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