Ta có
$a+b+c=3abc \Leftrightarrow \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=3\Leftrightarrow xy+yz+zx=3(x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c})$ Mặt khác
$\frac{bc}{a^{3}(c+2b)}=\frac{1}{a^{3}(\frac{c+2b}{bc})}=\frac{1}{a^{3}(\frac{1}{b}+\frac{2}{c})}=\frac{x^{3}}{y+2z}=\frac{x^{4}}{xy+2zx}$
$\Rightarrow Bt=\sum\frac{x^{4}}{xy+2zx}\geqslant \frac{(x^{2}+y^{2}+z^{2})^{2}}{3(xy+yz+zx)}\geqslant \frac{(xy+yz+zx)^{2}}{3(xy+yz+zx) }\geqslant \frac{xy+yz+zx}{3}=1$