Dự đoán của chúa Pain  x=y=z=1/3

áp dụng bất đẳng thức cô si ta có:

\(2xy\le2\left(\frac{x+y}{2}\right)^2\)

\(yz\le\left(\frac{y+z}{2}\right)^2\)

\(xz\le\left(\frac{z+x}{2}\right)^2\)

 ( vì X=Y=Z dự đoán của chúa pain) suy ra x+y=2x..ta được :

\(P\le2\left(\frac{x+y}{2}\right)^2+\left(\frac{y+z}{2}\right)^2+\left(\frac{z+x}{2}\right)^2\Leftrightarrow2x^2+y^2+z^2\)

\(P\le2x^2+y^2+z^2\Leftrightarrow P\le\frac{1}{3}\Leftrightarrow P\le\frac{2}{9}+\frac{1}{9}+\frac{1}{9}\Leftrightarrow P\le\frac{4}{9}\)

Vậy Max của P là 4/9 dâu = xảy ra khi x=y=z=1/3 đúng như dự đoán của chúa pain . chúa pain vô cmm nó địch :))

Cái chỗ \(P\le\frac{1}{3}\)

 là Mình viết nhầm nha 

Pain Thiên Đạo mới học lớp 8 nha bạn 

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

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

cô si cho gt

\(=>A=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{z-3}}{z}\)

áp dụng BĐT AM-GM

\(=>\sqrt{x-1}\le\dfrac{x-1+1}{2}=\dfrac{x}{2}\)

\(=>\dfrac{\sqrt{x-1}}{x}\le\dfrac{\dfrac{x}{2}}{x}=\dfrac{1}{2}\left(1\right)\)

có \(\dfrac{\sqrt{y-2}}{y}=\dfrac{\sqrt{\left(y-2\right)2}}{\sqrt{2}.y}\)

\(=>\sqrt{\left(y-2\right)2}\le\dfrac{y-2+2}{2}=\dfrac{y}{2}\)

\(=>\dfrac{\sqrt{\left(y-2\right)2}}{\sqrt{2}.y}\le\dfrac{\dfrac{y}{2}}{\sqrt{2}.y}=\dfrac{1}{2\sqrt{2}}\left(2\right)\)

tương tự \(=>\dfrac{\sqrt{z-3}}{z}\le\dfrac{1}{2\sqrt{3}}\left(3\right)\)

(1)(2)(3)\(=>A\le\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\)

Ta có: \(x+y+z=1\Rightarrow\hept{\begin{cases}\sqrt{x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\\\sqrt{y+xz}=\sqrt{y\left(x+y+z\right)+xz}=\sqrt{\left(x+y\right)\left(y+z\right)}\\\sqrt{z+xy}=\sqrt{z\left(x+y+z\right)+xy}=\sqrt{\left(x+z\right)\left(y+z\right)}\end{cases}}\)

Ta viết lại A

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

Áp dụng bđt AM-GM:

\(A\le\frac{x+y+x+z+x+y+y+z+y+z+x+z}{2}=2\)

\("="\Leftrightarrow x=y=z=\frac{1}{3}\)

\(x+yz=x\left(x+y+z\right)+yz\)

\(=x^2+xy+xz+yz\)

\(=x\left(x+y\right)+z\left(x+y\right)=\left(x+z\right)\left(x+y\right)\)

+ Tương tự : \(y+xz=\left(x+y\right)\left(y+z\right)\)

\(z+xy=\left(x+z\right)\left(y+z\right)\)

+ Theo bđt AM-GM : \(\sqrt{\left(x+y\right)\left(x+z\right)}\le\frac{x+y+x+z}{2}\)

\(\Rightarrow\sqrt{\left(x-1\right)\left(y-1\right)}\le\frac{2x+y+z}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x+y=x+z\Leftrightarrow y=z\)

+ Tương tự ta cm đc : 

\(\sqrt{\left(x+y\right)\left(y+z\right)}\le\frac{x+2y+z}{2}\).   Dấu "=" xảy ra \(\Leftrightarrow x=z\)

\(\sqrt{\left(x+z\right)\left(y+z\right)}\le\frac{x+y+2z}{2}\).   Dấu "=" xảy ra \(\Leftrightarrow x=y\)

Do đó : \(A\le\frac{4\left(x+y+z\right)}{2}=2\)

A = 2 \(\Leftrightarrow x=y=z=\frac{1}{3}\)

Vậy Max A = 2 \(\Leftrightarrow x=y=z=\frac{1}{3}\)

x+yz=x(x+y+z)+yz

=x2+xy+xz+yz

=x(x+y)+z(x+y)=(x+z)(x+y)

+ Tương tự : y+xz=(x+y)(y+z)

z+xy=(x+z)(y+z)

+ Theo bđt AM-GM : √(x+y)(x+z)≤x+y+x+z2 

⇒√(x−1)(y−1)≤2x+y+z2 

Dấu "=" xảy ra ⇔x+y=x+z⇔y=z

+ Tương tự ta cm đc : 

√(x+y)(y+z)≤x+2y+z2 .   Dấu "=" xảy ra ⇔x=z

√(x+z)(y+z)≤x+y+2z2 .   Dấu "=" xảy ra ⇔x=y

Do đó : A≤4(x+y+z)2 =2

A = 2 ⇔x=y=z=13 

Vậy Max A = 2