ÁD BĐT AM-GM,ta có:$(a+bc)(b+ca)\leq(\frac{a+bc+b+ca}{2})^{2}=\frac{(a+b)^{2}.(c+1)^{2}}{4}$
$TT$
$\Rightarrow\left[{(a+bc)(b+ca)(c+ab)}\right]^{2}\leq\left[ {(a+b)(b+c)(c+a)}\right]^{2}.\frac{(a+1)^{2}(b+1)^{2}(c+1)^{2}}{64}$
Ta cần CM:$\frac{(a+1)^{2}(b+1)^{2}(c+1)^{2}}{64}\leq1(1)$
Thật vậy:
Ta có:$(a+1)(b+1)(c+1)\leq (\frac{a+1+b+1+c+1}{3})^{3}=8$
$\Rightarrow(1)$ luôn đúng
$\Rightarrow$đpcm
Dấu''='' xra$\Leftrightarrow a=b=c=1$