|
|
Đặt $a= \dfrac{1}{x}, b= \dfrac{1}{y},c= \dfrac{1}{z} \Rightarrow xyz=1.$
Ta có
$\dfrac{2}{a^3\left(b+c\right)}=\dfrac{2}{\dfrac{1}{x^3}\left(\dfrac{1}{y}+ \dfrac{1}{z}\right)}= \dfrac{2x^2}{y+z}$, do $xyz=1.$
Như vậy
Vế trái $=\sum\dfrac{2}{a^3\left(b+c\right)}=\sum\dfrac{2x^2}{y+z}$
Áp dụng AM-GM ta có
$\dfrac{2x^2}{y+z} + \dfrac{y+z}{2} \ge 2\sqrt{\dfrac{2x^2}{y+z} . \dfrac{y+z}{2} }=2x $
Suy ra
$\sum\dfrac{2x^2}{y+z}+\sum\dfrac{y+z}{2} \ge \sum2x$
$\Leftrightarrow \sum\dfrac{2x^2}{y+z}+x+y+z \ge 2(x+y+z)$
$\Leftrightarrow \sum\dfrac{2x^2}{y+z} \ge x+y+z $
$\Leftrightarrow \sum\dfrac{2x^2}{y+z} \ge x+y+z \ge 3\sqrt[3]{xyz}=3$
Vậy
$\sum\dfrac{2}{a^3\left(b+c\right)}\ge 3$, đpcm.
Dấu bằng xảy ra khi $a=b=c=1.$
|