Giải thích các bước giải:
Ta có:
$A=2(x+y+z)+3(\dfrac1x+\dfrac1y+\dfrac1z)$
$\to A\ge 2(x+y+z)+3\cdot\dfrac{9}{x+y+z}$
$\to A\ge 2(x+y+z)+\dfrac{27}{x+y+z}$
$\to A\ge 2((x+y+z)+\dfrac{1}{x+y+z})+\dfrac{25}{x+y+z}$
$\to A\ge 2\cdot 2\sqrt{(x+y+z)\cdot\dfrac{1}{x+y+z}}+\dfrac{25}{1}$
$\to A\ge 29$
Dấu = xảy ra khi $x=y=z=\dfrac13$