Đáp án:

Đặt $\left\{\begin{matrix} a=\dfrac{1}{x} & \\ b=\dfrac{1}{y} & \\ c=\dfrac{1}{z}& \end{matrix}\right.$

Vì $a+b+c=6abc$

$→\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{6}{xyz}$

$→\dfrac{xy+yz+xz}{xyz}=\dfrac{6}{xyz}$

$\to xy+yz+xz=6$

$\dfrac{bc}{a^3(c+2b)}=\dfrac{\dfrac{1}{yz}}{\dfrac{1}{x^3z}+\dfrac{2}{x^3y}}= \dfrac{x^3}{y+2z}$

Tương tự, ta có: $\dfrac{ca}{b^3(a+2c)}=\dfrac{y^3}{z+2x}; \dfrac{ab}{c^3(b+2a)}=\dfrac{z^3}{x+2y}$

$→$ Bất đẳng thức cần chứng minh trở thành:

$\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}\geq 2$

\[\to VT=\dfrac{x^4}{yx+2zx}\geq \dfrac{(x^2+y^2+z^2)^2}{3(xy+yz+zx)}\]

mà $x^2+y^2+z^2\ge xy+yz+xz$

\[\to \dfrac{(x^2+y^2+z^2)^2}{3(xy+yz+zx)} \geq \dfrac{(xy+yz+zx)^2}{3(xy+yz+zx)}=2\]

\[\to \dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}\geq 2\]

\[\to \dfrac{bc}{a^{3}(c+2b)}+\dfrac{ca}{b^{3}(a+2c)}+\dfrac{ab}{c^{3}(b+2a)}\geq 2\]

Dấu $"="$ xảy ra $\Leftrightarrow x=y=z=\sqrt{2}\Leftrightarrow a=b=c=\dfrac{1}{\sqrt{2}}$