H
huy14112
Đội 4
Câu 1 :
$2(xy+yz+xz) \le \dfrac{2}{3}.(x+y+z)^2 = \dfrac{2}{3}$
$mxyz \le m. \dfrac{(x+y+z)^3}{27}= \dfrac{m}{27}$
$\rightarrow A \le \dfrac{2}{3} + \dfrac{m}{27}$
Sao lại có m nhỉ ???
Câu 1 :
$2(xy+yz+xz) \le \dfrac{2}{3}.(x+y+z)^2 = \dfrac{2}{3}$
$mxyz \le m. \dfrac{(x+y+z)^3}{27}= \dfrac{m}{27}$
$\rightarrow A \le \dfrac{2}{3} + \dfrac{m}{27}$
Sao lại có m nhỉ ???