\(2x+2y+z=4\Rightarrow z=4-2x-2y\)

Ta có: \(A=2xy+yz+xz\)

               \(=2xy+y\left(4-2x-2y\right)+x\left(4-2x-2y\right)\)

               \(=2xy+4y-2xy-2y^2+4x-2x^2-2xy\)

               \(=4y-2xy-2y^2+4x-2x^2\)

  \(\Rightarrow2A=8y-4xy-4y^2+8x-4x^2\)

               \(=-4x^2-4x\left(y-2\right)-4y^2+8y\)

               \(=-4x^2-2.x.2\left(y-2\right)-\left(y-2\right)^2+\left(y-2\right)^2-4y^2+8y\)

               \(=-\left[4x^2+2.x.2\left(y-2\right)+\left(y-2\right)^2\right]+\left(y-2\right)^2-4y^2+8y\)

                 \(=-\left(2x+y-2\right)^2+y^2-4y+4-4x^2+8y\)

                   \(=-\left(2x+y-2\right)^2-3y^2+4y+4\)        

                     \(=-\left(2x+y-2\right)^2-3\left(y^2-2.\frac{2}{3}y+\frac{4}{9}-\frac{4}{9}-\frac{4}{3}\right)\)       

                      \(=-\left(2x+y-2\right)^2-3\left(y-\frac{2}{3}\right)^2+\frac{16}{3}\)

                        \(=\frac{16}{3}-\left[\left(2x+y-2\right)^2+3\left(y-\frac{2}{3}\right)^2\right]\)

Vì \(\left(2x+y-2\right)^2\ge0;\left(y-\frac{2}{3}\right)^2\ge0\) Nên \(\frac{16}{3}-\left[\left(2x+y-2\right)^2+3\left(y-\frac{2}{3}\right)^2\right]\le\frac{16}{3}\)

\(\Rightarrow A\le\frac{16}{3}:2=\frac{8}{3}\)

Dấu "=" xảy ra <=>\(\hept{\begin{cases}y-\frac{2}{3}=0\\2x+y-2=0\\z=4-2x-2y\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-y+2}{2}\\y=\frac{2}{3}\\z=4-2x-2y\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=\frac{2}{3}\\z=\frac{4}{3}\end{cases}}}\)

Vậy A_Max = 8/3 khi và chỉ khi x = y = 2/3 và z = 4/3

\(P=\sqrt{x\left(x+y+z\right)+yz}+\sqrt{y\left(x+y+z\right)+xz}+\sqrt{z\left(x+y+z\right)+xy}\)

\(P=\sqrt{\left(x+y\right)\left(x+z\right)}+\sqrt{\left(x+y\right)\left(y+z\right)}+\sqrt{\left(x+z\right)\left(y+z\right)}\)

\(P\le\dfrac{1}{2}\left(x+y+x+z\right)+\dfrac{1}{2}\left(x+y+y+z\right)+\dfrac{1}{2}\left(x+z+y+z\right)\)

\(P\le2\left(x+y+z\right)=2\)

\(P_{max}=2\) khi \(x=y=z=\dfrac{1}{3}\)

\(A=2xy+yz+xz\)

\(=2xy+y\left(4-2x-2y\right)+x\left(4-2x-2y\right)\)

\(=-2x^2-2xy+4x-2y^2+4y\)

\(=\left[-\left(x^2+2xy+y^2\right)+\dfrac{8}{3}\left(x+y\right)-\dfrac{16}{9}\right]-\left(x^2-\dfrac{4}{3}x+\dfrac{4}{9}\right)-\left(y-\dfrac{4}{3}y+\dfrac{4}{9}\right)+\dfrac{8}{3}\)\(=-\left(x+y-\dfrac{4}{3}\right)^2-\left(x-\dfrac{2}{3}\right)^2-\left(y-\dfrac{2}{3}\right)^2+\dfrac{8}{3}\le\dfrac{8}{3}\)

Vậy \(A_{max}=\dfrac{8}{3}\) tại \(\left\{{}\begin{matrix}x=y=\dfrac{2}{3}\\z=\dfrac{4}{3}\end{matrix}\right.\)

z = 4-2(x+y)

=> A= 2xy + y[4-2(x+y)] + x[4-2(x+y)]

=\(2xy+4y-2xy-2y^2+4x-2x^2-2xy\)

= \(-\left(y^2-4y+4\right)-\left(x^2-4x+4\right)-\left(y^2+2xy+x^2\right)+8\)

=\(8-\left[\left(y-2\right)^2+\left(x-2\right)^2-\left(y-x\right)^2\right]\le8\forall x,y\)

vậy GTLN của A là 8 khi x=y=2

\(x+\sqrt{x+yz}=x+\sqrt{x\left(x+y+z\right)+yz}=x+\sqrt{x^2+yz+x\left(z+y\right)}\)

\(\ge x+\sqrt{2\sqrt{x^2yz}+x\left(y+z\right)}=x+\sqrt{x\cdot2\sqrt{yz}+x\left(y+z\right)}=x+\sqrt{x\left(y+z+2\sqrt{yz}\right)}\)

\(=x+\sqrt{x\left(\sqrt{y}+\sqrt{z}\right)^2}=x+\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)\)

\(\Rightarrow\frac{x}{x+\sqrt{x+yz}}\le\frac{x}{x+\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)}=\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

tương tự :

\(\frac{y}{y+\sqrt{y+xz}}\le\frac{\sqrt{y}}{\sqrt{y}+\sqrt{x}+\sqrt{z}}\)

\(\frac{z}{z+\sqrt{z+xy}}\le\frac{\sqrt{z}}{\sqrt{z}+\sqrt{x}+\sqrt{y}}\) 

cộng vế theo vế ta được 

\(\frac{x}{x+\sqrt{x+yz}}+\frac{y}{y+\sqrt{y+zx}}+\frac{z}{z+\sqrt{z+xy}}\le\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)

dấu "=" xảy tra khi x=y=z=1/3

cái này thì chịu

khó muốn chết luôn làm sao làm đc

cô si cho gt

\(xy+yz+zx\le\dfrac{1}{3}\left(x+y+z\right)^2=\dfrac{4}{3}\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{2}{3}\)