Câu 4: Ta có:
$12x^2+12y^2+3z^2=\frac{x^2}{\frac{1}{12}}+\frac{y^2}{\frac{1}{12}}+\frac{z^2}{\frac{1}{3}}$
$\geq \frac{(x+y+z)^2}{\frac{1}{12}+\frac{1}{12}+\frac{1}{3}}=2(x+y+z)^2$
Suy ra: $A\geq 4(xy+yz+zx)=4$
Dấu bằng xảy ra khi: $x=y=\frac{1}{3};z=\frac{4}{3}$
- BachMieu yêu thích