[Đại 9] Chứng minh bất đẳng thức

P

pl09

Last edited by a moderator:
V

vipboycodon

Theo bdt cauchy - schwarz ta có:
$\dfrac{4}{2a+b+c}+\dfrac{4}{2b+a+c}+\dfrac{4}{2c+a+b} \ge \dfrac{(2+2+2)^2}{2a+b+c+2b+a+c+2c+a+b} = \dfrac{36}{4(a+b+c)} = \dfrac{9}{a+b+c}$
Dấu "=" xảy ra khi $\dfrac{2}{2a+b+c} = \dfrac{2}{2b+a+c} = \dfrac{2}{2c+a+b}$ <=> $a = b = c$
 
Last edited by a moderator:
You must log in or register to reply here.
Top Bottom