CM BĐT phụ:
[tex](xy+yz+xz)^2\geq 3xyz(x+y+z)\\[/tex]
[tex]x+y+z\geq 3\sqrt[3]{xyz}\Rightarrow xyz\leq \frac{1}{27}[/tex]
Khi đó:
[tex]A\geq 9xyz-12xyz=-3xyz\\[/tex] [tex]\geq -3.\frac{1}{27}=\frac{-1}{9}[/tex]
Dấu "=" xảy ra <=> x = y = z = 1/3
CM BĐT phụ:
[tex](xy+yz+xz)^2\geq 3xyz(x+y+z)\\[/tex]
[tex]x+y+z\geq 3\sqrt[3]{xyz}\Rightarrow xyz\leq \frac{1}{27}[/tex]
Khi đó:
[tex]A\geq 9xyz-12xyz=-3xyz\\[/tex] [tex]\geq -3.\frac{1}{27}=\frac{-1}{9}[/tex]
Dấu "=" xảy ra <=> x = y = z = 1/3