proving for real positive numbers

**nh149** · July 22nd 2009, 11:10 PM

how can i do?
For a,b,c real positive numbers, show that ;

(a+b+c)(1/a + 1/b + 1/c)>= 9

is there any theorem to use?
i dont know theorems based on this..
can u help me with this?

**alexmahone** · July 22nd 2009, 11:46 PM

Using the AM-GM-HM inequality,

$AM = \frac{a+b+c}{3}$

$HM = \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$

$AM \geq GM \geq HM$

$AM \geq HM$

$\frac{a+b+c}{3} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$

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

**nh149** · July 23rd 2009, 12:15 AM

i don't even know the fact that

.
thankx for ur help alexmahone..

i really appreciate it..

**BobP** · July 23rd 2009, 12:34 AM

You could start with
$(a - b )^2 \geq 0, \mbox{ so } a^2 + b^2 - 2ab \geq 0$ .
Then,
$\ a^2 + b^2 \geq 2ab \mbox{ so } \frac{a^2 + b^2}{ab} \geq 2 \mbox{ ( given that } a \mbox{ and } b$ are positive numbers).
So, multiply out the LHS of your inequality and collect up terms. The two terms involving $a \mbox{ and } b$ can be combined to produce the fraction $( a^2 + b^2 ) / ab$ , so you can make use of the result obtained earlier.
Do the same for the terms involving $b \mbox{ and } c$ and those involving $c \mbox{ and } a$ , and the result follows.

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proving for real positive numbers

thank u very much..

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