Bài 1:
Ta có : $x,y,z\geq 0$
$\Rightarrow x+y+ z\geq 3\sqrt[3]{xyz}$ $( "=" khi x=y=z=\frac{1}{3})$
$\Rightarrow \sqrt[3]{xyz}\leq \frac{1}{3}$
$\Rightarrow xyz\leq \frac{1}{27} (1)$
Lại có:
$(x+y)+(y+z)+(z+x) \geq 3\sqrt[3]{(x+y)(y+z)(z+x)}$
$\sqrt[3]{(x+y)(y+z)(z+x)}\leq \frac{2}{3}$
$\Rightarrow (x+y)(y+z)(z+x)\leq \frac{8}{27} (2)$
$Từ (1) và (2) \Rightarrow A \leq \frac{8}{27^{2}}$
Vậy$ Max A=\frac{8}{729} khi x=y=z=\frac{1}{3}$