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

Mathematics
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sure its difficult ?

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Other answers:

Unless you know the first step it is difficult. :)
i suck at these
No worries, either the viewers will let you out from this question or I will at the end.
Here are many legends viewing this, I would like a better answer from them.
if x >0, then:\[x+\frac{1}{x}\ge2\]you can use this to show the desired inequality.
hello joe!!!
hey :)
i was thinking harmonic mean and geometric mean or something but i bet i was wrong
long time, how you been?
not bad, its senior year, so work work work lol
@joemath314159 how? I am curious to know that how will it work?
keep busy, glad to see you!
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.
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.
AM>=HM (a+b+c)/3 >= 3/(1/a+1/b+1/c)
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) \[\large{\frac{a+b+c}{3} \times \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3} \ge (abc)^{\frac{1}{3}} \times (\frac{1}{abc})^{\frac{1}{3}}}\] \[\large{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge 9}\]
And.,.... Hence proved.. Right @joemath314159 ?
very interesting :)
:) yes .
finally u are doing only AM>=HM only
but other way round......
yep :)

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