Ta có :
$$x^3+\left(\frac{2-\sqrt 2}2\right)^3+\left(\frac{2-\sqrt 2}2\right)^3 \ge 3\left(\frac{2-\sqrt 2}2\right)^2x$$
$$y^3+\left(\frac{2-\sqrt 2}2\right)^3+\left(\frac{2-\sqrt 2}2\right)^3 \ge 3\left(\frac{2-\sqrt 2}2\right)^2y$$
$$\frac {z^3}2+\sqrt 2\left(\frac{2-\sqrt 2}2\right)^3+\sqrt 2\left(\frac{2-\sqrt 2}2\right)^3 \ge 3\left(\frac{2-\sqrt 2}2\right)^2x$$
$$\Rightarrow x^3+y^3+\frac{z^3}3+(4+2\sqrt 2)\left(\frac{2-\sqrt 2}2\right)^3 \ge 3\left(\frac{2-\sqrt 2}2\right)^2$$
$$\Rightarrow A \ge \frac{3-2\sqrt 2}{2}$$
Dấu $"="$ xảy ra$\Leftrightarrow x=y= \frac{2-\sqrt 2}2, z=\sqrt 2-1$