## mathslover 3 years ago if : a > 0 , b > 0 and c>0 , prove that: $$\large{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge 9}$$

1. mathslover

@hartnn

2. mathslover

@Zarkon and @myininaya

3. hartnn

sure its difficult ?

4. mathslover

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

5. satellite73

i suck at these

6. mathslover

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

7. mathslover

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

8. joemath314159

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

9. satellite73

hello joe!!!

10. joemath314159

hey :)

11. satellite73

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

12. satellite73

long time, how you been?

13. joemath314159

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

14. mathslover

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

15. satellite73

keep busy, glad to see you!

16. joemath314159

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.

17. joemath314159

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.

18. hartnn

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

19. mathslover

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

20. mathslover

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

21. joemath314159

very interesting :)

22. mathslover

:) yes .

23. hartnn

finally u are doing only AM>=HM only

24. hartnn

but other way round......

25. mathslover

yep :)