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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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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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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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