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?
Using the AM-GM-HM inequality,
$\displaystyle AM = \frac{a+b+c}{3} $
$\displaystyle HM = \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$
$\displaystyle AM \geq GM \geq HM$
$\displaystyle AM \geq HM$
$\displaystyle \frac{a+b+c}{3} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} $
$\displaystyle (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq 9$
You could start with
$\displaystyle (a - b )^2 \geq 0, \mbox{ so } a^2 + b^2 - 2ab \geq 0 $.
Then,
$\displaystyle \ 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 $\displaystyle a \mbox{ and } b $ can be combined to produce the fraction $\displaystyle ( a^2 + b^2 ) / ab $, so you can make use of the result obtained earlier.
Do the same for the terms involving $\displaystyle b \mbox{ and } c $ and those involving $\displaystyle c \mbox{ and } a $ , and the result follows.