A community for students.
Here's the question you clicked on:
 0 viewing
mathslover
 2 years ago
if : a > 0 , b > 0 and c>0 , prove that:
\(\large{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge 9}\)
mathslover
 2 years ago
if : a > 0 , b > 0 and c>0 , prove that: \(\large{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge 9}\)

This Question is Closed

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.2@Zarkon and @myininaya

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.2Unless you know the first step it is difficult. :)

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

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

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.2if x >0, then:\[x+\frac{1}{x}\ge2\]you can use this to show the desired inequality.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1i was thinking harmonic mean and geometric mean or something but i bet i was wrong

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1long time, how you been?

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.2not bad, its senior year, so work work work lol

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1keep busy, glad to see you!

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.2So \[(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=3+(\frac{a}{b}+\frac{b}{a})+(\frac{a}{c}+\frac{c}{a})+(\frac{c}{b}+\frac{b}{c})\]Since a,b,c are all positive, the fractions are all positive. That inequality i posted applies to all three parenthesis.

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.2if\[x=\frac{a}{b}\]then\[\frac{a}{b}+\frac{b}{a}=x+\frac{1}{x}\]which is greater than 2. Similarly for the other guys.

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.0AM>=HM (a+b+c)/3 >= 3/(1/a+1/b+1/c)

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

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.2And.,.... Hence proved.. Right @joemath314159 ?

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.2very interesting :)

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.0finally u are doing only AM>=HM only

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.0but other way round......
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.