Bđt

V

vipboycodon

$(\sum \dfrac{ab}{c})^2 = \sum \dfrac{a^2b^2}{c^2}+ 2 \sum a^2$
áp dụng cô-si ta có:
$\dfrac{a^2b^2}{c^2}+\dfrac{a^2c^2}{b^2} \ge 2a^2$
$\dfrac{a^2b^2}{c^2}+\dfrac{b^2c^2}{a^2} \ge 2b^2$
$\dfrac{b^2c^2}{a^2}+\dfrac{a^2c^2}{b^2} \ge 2c^2$
=> $\sum \dfrac{a^2b^2}{c^2} \ge \sum a^2$
=> $(\sum \dfrac{ab}{c})^2 \ge 3 \sum a^2 = 9$
=> đpcm
Dấu "=" xảy ra khi a = b = c = 1
 
Last edited by a moderator:
L

lp_qt

đặt $\left\{\begin{matrix}x=\dfrac{ab}{c} & \\ y=\dfrac{bc}{a} & \\ z=\dfrac{ac}{b} &
\end{matrix}\right. (x;y;z >0)$

\Rightarrow $\left\{\begin{matrix}b^2=xy & \\ a^2=xz & \\ c^2=yz & \end{matrix}\right.$

\Rightarrow $xy+yz+xy=a^2+b^2+c^2=3$

ta có

$(x+y+z)^2\ge 3.(xy+yz+xz)=3.3=9$

\Rightarrow $x+y+z \ge 3$
\Rightarrow đpcm
 
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