Giải thích các bước giải:
$18\ge x(x+1)+y(y+1)+z(z+1)=x^2+y^2+z^2+(x+y+z)\ge \dfrac{(x+y+z)^2}{3} +(x+y+z)$
$\rightarrow (x+y+z)^2+3(x+y+z)-54\le 0$
$\rightarrow (x+y+z-6)(x+y+z+9)\le 0$
$\rightarrow x+y+z-6\le 0$
$\rightarrow x+y+z\le 6$
Lại có :
$B=\dfrac{1}{x+y+1}+\dfrac{1}{y+z+1}+\dfrac{1}{z+x+1}$
$\rightarrow B\ge\dfrac{9}{x+y+1+y+z+1+z+x+1}$
$\rightarrow B\ge\dfrac{9}{2(x+y+z)+3}$
$\rightarrow B\ge\dfrac{9}{2.6+3}=\dfrac{3}{5}$