Cho 3 số thực (x,y,z) thỏa mãn (2x + 2y + z = 4.) Tìm giá trị lớn nhất của biểu thức: (A = 2xy + yz + zx.)

Ta có: (2x + 2y + z = 4 Leftrightarrow z = 4 – 2x – 2y.)

(begin{array}{l} Rightarrow A = 2xy + yz + zx = 2xy + yleft( {4-2x-2y} right) + xleft( {4-2x-2y} right)\ = {rm{ }}2xy + 4y-2xy-2{y^2} + 4x-2{x^2}-2xy\ = -2{x^2}-2xy + 4x-2{y^2} + 4y\ =  – left[ {left( {{x^2} + 2xy + {y^2}} right) – frac{8}{3}left( {x + y} right) + frac{{16}}{9}} right] – left( {{x^2} – frac{4}{3}x + frac{4}{9}} right) – left( {{y^2} – frac{4}{3}y + frac{4}{9}} right) + frac{8}{3}\ =  – {left( {x + y – frac{4}{3}} right)^2} – {left( {x – frac{2}{3}} right)^2} – {left( {y – frac{2}{3}} right)^2} + frac{8}{3}end{array})         

Mà: ({left( {x + y – frac{4}{3}} right)^2} ge 0;;{left( {x – frac{2}{3}} right)^2} ge 0;;{left( {y – frac{2}{3}} right)^2} ge 0;forall x,;y)

( Rightarrow A =  – {left( {x + y – frac{4}{3}} right)^2} – {left( {x – frac{2}{3}} right)^2} – {left( {y – frac{2}{3}} right)^2} + frac{8}{3} ge frac{8}{3})(forall x,;y)

Dấu “=” xảy ra ( Leftrightarrow left{ begin{array}{l}x + y – frac{4}{3} = 0\x – frac{2}{3} = 0\y – frac{2}{3} = 0end{array} right. Leftrightarrow x = y = frac{2}{3}.)

Vậy Amax = (frac{8}{3}) tại  (left{ begin{array}{l}x = y = frac{2}{3}\z = frac{4}{3}end{array} right.).  

Chọn D.