\(x^2y^2+\left(x-2\right)^2+\left(2y-2\right)^2-2xy\left(x+2y-4\right)=0\)

<=> \(x^2y^2+\left(x+2y-4\right)^2-2\left(x-2\right)\left(2y-2\right)-2xy\left(x+2y-4\right)=0\)

<=> \(\left[x^2y^2-2xy\left(x+2y-4\right)+\left(x+2y-4\right)^2\right]-4\left(xy-x-2y+2\right)=0\)

<=> \(\left(xy-x-2y+4\right)^2-4\left(xy-x-2y+4\right)+8=0\)

<=> \(\left(xy-x-2y+2\right)^2+4=0\)(vô nghiệm)

=>phương trình vô nghiệm

                                                                     Giải

5 = x²y² + ( x-2) ² + ( 2y-2)² -2xy(x + 2y -4 )

    = [ x.y - ( x + 2.y -4 ) ] ² - 2 ( y - 1 ) ( x - 2  ) 

    = ( xy - x - 2y + 4 )² -4.( xy - x - 2y + 2 )

    = A²  - 4 ( A - 2 )

    <=> A² - 4.A + 3 = 0

    <=>   \(\orbr{\begin{cases}xy-x-2y+4=3\\xy-x-2y+4=1\end{cases}}\)

Lưu ý : đặt : A = xy - x - 2y + 4 

TH1 : xy - x - 2.y + 4  = 3 

<=> xy - x - 2y + 1        = 0 

<=> x.( y  - 1 ) - 2.(y-1 ) = 1

<=> ( x - 2 )  (  y - 1 ) = 1 

Ta có bảng : 

TH2 : xy - x - 2y + 4 = 1 

<=> ( x- 2 ) . ( y -1 ) =-1 

helloo

Ta có \(1+x^2=x^2+xy+yz+xz=\left(x+z\right)\left(x+y\right)\)

Khi đó BĐT <=>

 \(\frac{1}{\left(x+y\right)\left(x+z\right)}+\frac{1}{\left(y+z\right)\left(x+z\right)}+\frac{1}{\left(x+y\right)\left(y+z\right)}\ge\frac{2}{3}\left(\frac{x}{\sqrt{\left(x+z\right)\left(x+y\right)}}+...\right)\)

<=> \(\frac{x+y+z}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\frac{x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}}{\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}}\right)^3\)

<=>\(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge\frac{1}{3}\left(x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}\right)^3\)

<=> \(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\right)^3\)(1)

Xét \(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge\frac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)

<=> \(9\left[xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\right]\ge8\left(xy\left(x+y\right)+xz\left(x+z\right)+yz\left(y+z\right)+3xyz\right)\)

<=> \(xy\left(y+x\right)+yz\left(y+z\right)+xz\left(x+z\right)\ge6xyz\)

<=> \(x\left(y-z\right)^2+z\left(x-y\right)^2+y\left(x-z\right)^2\ge0\)luôn đúng

Khi đó (1) <=> 

\(\left(x+y+z\right).\frac{2\sqrt{2}}{3}.\sqrt{x+y+z}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+....\right)^3\) 

<=> \(\sqrt{2\left(x+y+z\right)}\ge\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\)

Áp dụng buniacopxki cho vế phải ta có 

\(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\le\sqrt{\left(x+y+z\right)\left(3-xy-yz-xz\right)}\)

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

=> BĐT được CM

Dấu bằng xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)

a=-2 b xấp xỉ 5.4