how can i do

(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? (Lipssealed)

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- Jul 22nd 2009, 11:01 PMnh149proving
how can i do

(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? (Lipssealed) - Jul 22nd 2009, 11:05 PMpomp
- Jul 24th 2009, 10:03 AMJ.R
actually, we must just precise that a, b , c are three positif reals not null.

indeed, we can apply the Cauchy-Schwarz's inegalitie:

we have directly:

$\displaystyle (\sqrt{a}^2+\sqrt{b}^2+\sqrt{c}^2)((\frac{1}{\sqrt {a}})^2+(\frac{1}{\sqrt{b}})^2+(\frac{1}{\sqrt{c}} )^2)\ge (1+1+1)^2=9$

cqfd

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