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: \(2x + 2y + z = 4 \Leftrightarrow z = 4 - 2x - 2y.\)

\( \Rightarrow A = 2xy + yz + zx\)\( = 2xy + y\left( {4-2x-2y} \right) + x\left( {4-2x-2y} \right)\)\( = {\rm{ }}2xy + 4y-2xy-2{y^2} + 4x-2{x^2}-2xy\)\(= -2{x^2}-2xy + 4x-2{y^2} + 4y\)\( =  - \left[ {\left( {{x^2} + 2xy + {y^2}} \right) - \dfrac{8}{3}\left( {x + y} \right) + \dfrac{{16}}{9}} \right] - \left( {{x^2} - \dfrac{4}{3}x + \dfrac{4}{9}} \right) - \left( {{y^2} - \dfrac{4}{3}y + \dfrac{4}{9}} \right) + \dfrac{8}{3}\)\(=  - {\left( {x + y - \dfrac{4}{3}} \right)^2} - {\left( {x - \dfrac{2}{3}} \right)^2} - {\left( {y - \dfrac{2}{3}} \right)^2}+ \dfrac{8}{3}\)          

Mà: \({\left( {x + y - \dfrac{4}{3}} \right)^2} \ge 0;\;{\left( {x - \dfrac{2}{3}} \right)^2} \ge 0;\;{\left( {y - \dfrac{2}{3}} \right)^2} \ge 0\) với mọi \(x,\;y\)

\( \Rightarrow A =  - {\left( {x + y - \dfrac{4}{3}} \right)^2} - {\left( {x - \dfrac{2}{3}} \right)^2} - {\left( {y - \dfrac{2}{3}} \right)^2} + \dfrac{8}{3} \le \dfrac{8}{3}\) với mọi \( x,\;y\).

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

Vậy A_max = \(\dfrac{8}{3}\) khi  \(\left\{ \begin{array}{l}x = y = \dfrac{2}{3}\\z = \dfrac{4}{3}\end{array} \right.\).