if : a 0 , b 0 and c0 , prove that:

$\large{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge 9}$

Question

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 · Answer

@hartnn

mathslover · Answer

@Zarkon and @myininaya

hartnn · Answer

sure its difficult ?

mathslover · Answer

Unless you know the first step it is difficult. 
:)

anonymous · Answer

i suck at these

mathslover · Answer

No worries, either the viewers will let you out from this question or I will at the end.

mathslover · Answer

Here are many legends viewing this, I would like a better answer from them.

anonymous · Answer

if x >0, then:$$x+\frac{1}{x}\ge2$$you can use this to show the desired inequality.

anonymous · Answer

hello joe!!!

anonymous · Answer

hey :)

anonymous · Answer

i was thinking harmonic mean and geometric mean or something
but i bet i was wrong

anonymous · Answer

long time, how you been?

anonymous · Answer

not bad, its senior year, so work work work lol

mathslover · Answer

@joemath314159 how? I am curious to know that how will it work?

anonymous · Answer

keep busy, glad to see you!

anonymous · Answer

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.

anonymous · Answer

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.

hartnn · Answer

AM>=HM
(a+b+c)/3  >= 3/(1/a+1/b+1/c)

mathslover · Answer

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) 

$$\large{\frac{a+b+c}{3} \times \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3} \ge (abc)^{\frac{1}{3}} \times (\frac{1}{abc})^{\frac{1}{3}}}$$

$$\large{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge 9}$$

mathslover · Answer

And.,.... Hence proved.. 

Right @joemath314159 ?

anonymous · Answer

very interesting :)

mathslover · Answer

:) yes .

hartnn · Answer

finally u are doing only 
AM>=HM only

hartnn · Answer

but other way round......

mathslover · Answer

yep  :)