Tìm GTNN của biểu thức khó

V

viethoang1999

$a+b+c=abc$
\Leftrightarrow $\sum \dfrac{1}{ab}=1$
Ta có: $T=\dfrac{5}{a^2}+\dfrac{1}{b^2}+\dfrac{2}{c^2}$
$=\left ( \dfrac{3}{a^2}+\dfrac{1}{3b^2} \right ) +\left ( \dfrac{2}{3b^2}+\dfrac{3}{2c^2} \right ) +\left ( \dfrac{1}{2c^2}+\dfrac{2}{a^2} \right ) $
$\ge 2.\dfrac{1}{ab}+2.\dfrac{1}{bc}+2.\dfrac{1}{ca}=2 \sum \dfrac{1}{ab}=2$


 
You must log in or register to reply here.
Top Bottom