Các cao nhân giúp mình vs Cho x,y,z thoả x+y+z=1/3 Tìm GTLN của A=2xy+7yz+3xz

Đáp án: $A\le \dfrac7{30}$

Giải thích các bước giải:

 Ta có: $x+y+z=\dfrac13\to z=\dfrac13-x-y$

$\to A=2xy+7y\cdot \left(\dfrac13-x-y\right)+3x\cdot \left(\dfrac13-x-y\right)$

$\to A=-3x^2-8xy+x-7y^2+\dfrac{7y}{3}$

$\to -3A=-3\left(-3x^2-8xy+x-7y^2+\dfrac{7y}{3}\right)$

$\to -3A=9x^2+24xy-3x+21y^2-7y$

$\to -3A=9x^2+3x\left(8y-1\right)+21y^2-7y$

$\to -3A=\left(3x\right)^2+2\cdot 3x\cdot\dfrac{8y-1}{2}+\left(\dfrac{8y-1}{2}\right)^2-\left(\dfrac{8y-1}{2}\right)^2+21y^2-7y$

$\to -3A=\left(3x+\dfrac{8y-1}{2}\right)^2+\dfrac{20y^2-12y-1}{4}$

$\to -3A=\left(3x+\dfrac{8y-1}{2}\right)^2+\dfrac14\left(20\left(y-\dfrac{3}{10}\right)^2-\dfrac{14}{5}\right)$

$\to -3A=\left(3x+\dfrac{8y-1}{2}\right)^2+5\left(y-\dfrac{3}{10}\right)^2-\dfrac{7}{10}$

$\to -3A\ge 0+0-\dfrac{7}{10}$

$\to -3A\ge -\dfrac{7}{10}$

$\to A\le \dfrac7{30}$

Dấu = xảy ra khi:

$\begin{cases}y-\dfrac{3}{10}=0\\3x+\dfrac{8y-1}{2}=0\\ z=\dfrac13-x-y\end{cases}$

$\begin{cases}y=\dfrac{3}{10}\\x=-\dfrac7{30}\\ z=\dfrac{4}{15}\end{cases}$