Đặt x=$\frac{1}{a}$;y=$\frac{1}{2b}$;z=$\frac{1}{3c}$($x,y,z$>0)gt$\Leftrightarrow$$xyz$=1
P=$\frac{1}{a^{4}(2b+1)(3c+1)}$+$\frac{1}{16b^{4}(3c+1)(a+1)}$+$\frac{1}{81c^{4}(a+1)(2b+1)}$
$\Rightarrow$P=$\frac{x^{3}}{(y+1)(z+1)}$+$\frac{y^{3}}{(z+1)(x+1)}$+$\frac{z^{3}}{(x+1)(y+1)}$
Áp dụng BĐT Cauchy:
$\frac{x^{3}}{(y+1)(z+1)}$+$\frac{y+1}{8}$+$\frac{z+1}{8}$$\geq$$\frac{3x}{4}$
Tương tự$\Rightarrow$P$\geq$$\frac{1}{2}$($x+y+z$)-$\frac{3}{4}$$\geq$$\frac{1}{2}$.3$\sqrt[3]{xyz}$-$\frac{3}{4}$=$\frac{3}{4}$
Dấu''='' xra$\Leftrightarrow$a=1;b=$\frac{1}{2}$;c=$\frac{1}{3}$