Đặt $a=x^{3}$
$b=y^{3}$
$c=z^{3}$
Ta có: $a.b.c=1⇒x^{3}.y^{3}.z^{3}=1⇔x.y.z=1$
Ta có: $a+b=x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})$
Áp dụng bất đẳng thức Cô-si, ta có:
$a+b≥(x+y)(2xy-xy)=(x+y).xy$
Tương tự ta có: $b+c≥(y+z).yz$
$a+c≥(x+z).xz$
Thay vào $A$ ta có:
$A=\dfrac{1}{a+b+1}+\dfrac{1}{b+c+1}+\dfrac{1}{c+a+1}$
$⇔A≥\dfrac{1}{(x+y).xy+1}+\dfrac{1}{(y+z).yz+1}+\dfrac{1}{(x+z).xz+1}$
$⇔A≥\dfrac{z}{(x+y).xyz+z}+\dfrac{x}{(y+z).xyz+x}+\dfrac{y}{(x+z).xyz+y}$
$⇔A≥\dfrac{z}{x+y+z}+\dfrac{x}{x+y+z}+\dfrac{y}{x+y+z}$ $(x.y.z=1)$
$⇔A≥\dfrac{x+y+z}{x+y+z}=1$
Dấu $=$ xảy ra khi $x=y=z$ hay $a=b=c$
Vậy $A_{Min}=1$ khi $a=b=c$
