x+y+z=0

=>x+y=-z; x+z=-y; y+z=-x

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

=>\(x^2+2xy+y^2=z^2\)

=>\(z^2-xy=x^2+xy+y^2\)

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

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

\(xy+2z^2=xy+2(x^2+2xy+y^2)\)

\(=2x^2+5xy+2y^2=(x+2y)(2x+y)\)

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

=>\(x^2+2xz+z^2=y^2\)

=>\(y^2-xz=x^2+xz+z^2\)

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

\(=4yz-\left(y^2+2yz+z^2\right)=-y^2+2yz-z^2=-\left(y-z\right)^2\)

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

\(=2y^2+5yz+2z^2=2y^2+4yz+yz+2z^2\)

=2y(y+2z)+z(y+2z)

=(y+2z)(2y+z)

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

=>\(y^2+2yz+z^2=x^2\)

=>\(x^2-yz=y^2+yz+z^2\)

\(4xz-y^2\) =4xz-(x+z)^2

=4xz-\(x^2-2xz-z^2\)

\(=-x^2+2xz-z^2=-\left(x-z\right)^2\)

\(xz+2y^2=xz+2\left(x+z\right)^2\)

\(=xz+2x^2+4xz_{}+2z^2=2x^2+5xz+2z^2\)

\(=2x^2+4xz+xz+2z^2\)

=2x(x+2z)+z(x+2z)

=(x+2z)(2x+z)

2x+y=x+x+y=x-z

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

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

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

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

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

Ta có: \(A = \frac{4xy - z^2}{xy + 2z^2} \cdot \frac{4yz - x^2}{yz + 2x^2} \cdot \frac{4zx - y^2}{xz + 2y^2}\)

\(=\frac{-(x - y)^2}{(x + 2y)(2x + y)}\cdot\frac{-(y - z)^2}{(y + 2z)(2y + z)}\cdot\frac{-(z - x)^2}{(z + 2x)(2z + x)}\)

\(=\frac{-\left(x-y\right)^2\cdot\left(y-z\right)^2\cdot\left(z-x\right)^2}{(y-z)(x-z)(z-x)[-(x-y)](x-y)[-(y-z)]}\)

\(=\frac{-(x - y)^2 (y - z)^2 (z - x)^2}{-(x - y)^2 (y - z)^2 (z - x)^2}=1\)

Xem lại cái đề đi Tuyển. Hình như giá trị nhỏ nhất của cái biểu thức dưới còn lớn hơn là 1 thì làm sao bài đó có giá trị x, y, z thỏa được mà bảo tính A.

\(gt\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)

\(P=\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)

\(=\dfrac{1}{xyz}\left(x\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}+y\sqrt{\dfrac{5}{4}\left(x+z\right)^2+\dfrac{3}{4}\left(x-z\right)^2}+z\sqrt{\dfrac{5}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2}\right)\)

\(\ge\dfrac{1}{xyz}\left[x.\dfrac{\sqrt{5}\left(z+y\right)}{2}+y.\dfrac{\sqrt{5}\left(x+z\right)}{2}+z.\dfrac{\sqrt{5}\left(x+y\right)}{2}\right]\)

\(=\dfrac{\sqrt{5}\left(z+y\right)}{2yz}+\dfrac{\sqrt{5}\left(x+z\right)}{2xz}+\dfrac{\sqrt{5}\left(x+y\right)}{2xy}\)

\(=\dfrac{\sqrt{5}}{3}\left(1+1+1\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{\sqrt{5}}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2=\dfrac{\sqrt{5}}{3}\) (bunhia)

Dấu = xảy ra khi \(x=y=z=9\)

 Thấy : \(\sqrt{2y^2+yz+2z^2}=\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)>0\) 

CMTT : \(\sqrt{2x^2+xz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\)  ; \(\sqrt{2y^2+xy+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\) 

Suy ra : \(P\ge\dfrac{1}{xyz}.\dfrac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]\)

\(\Rightarrow P\ge\sqrt{5}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) 

Ta có : \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=\sqrt{xyz}\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\) 

Mặt khác :   \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2}{3}=\dfrac{1}{3}\)

Suy ra : \(P\ge\dfrac{\sqrt{5}}{3}\)

" = " \(\Leftrightarrow x=y=z=9\)

\(VT=6\left(x^2+y^2+z^2\right)+10\left(xy+yz+xz\right)+2\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)

\(=6\left(x+y+z\right)^2-2\left(xy+yz+xz\right)+2\frac{9}{2x+y+z+x+2y+z+x+y+2z}\)

\(\ge6\left(x+y+z\right)^2-2\frac{\left(x+y+z\right)^2}{3}+2\frac{9}{4\left(x+y+z\right)}\)

\(=\: 6\cdot\left(\frac{3}{4}\right)^2-2\cdot\frac{\left(\frac{3}{4}\right)^2}{3}+2\cdot\frac{9}{4\cdot\frac{3}{4}}=9\)

\(M=\dfrac{xy+2x+1}{xy+x+y+1}+\dfrac{yz+2y+1}{yz+y+z+1}+\dfrac{xz+2z+1}{xz+z+x+1}\)

\(M=\dfrac{xy+x+x+1}{x\left(y+1\right)+y+1}+\dfrac{yz+y+y+1}{y\left(z+1\right)+z+1}+\dfrac{xz+z+z+1}{z\left(x+1\right)+x+1}\)

\(\Rightarrow M=\dfrac{x\left(y+1\right)+x+1}{\left(x+1\right)\left(y+1\right)}+\dfrac{y\left(z+1\right)+y+1}{\left(y+1\right)\left(z+1\right)}+\dfrac{z\left(x+1\right)+z+1}{\left(z+1\right)\left(x+1\right)}\)

Quy đồng là xong nha

Ta có:

\(xy+yz+zx=4xyz\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\)

\(P=\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\)

\(\le\frac{1}{16}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}+\frac{2}{y}+\frac{1}{z}+\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\right)\)

\(\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)

áp dụng cô si sháp cho 4 số ta được :

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{16}{a+b+c+d}\)  Luôn đúng , ( tự chứng minh )

\(\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\ge\frac{1}{a+b+c+d}\) luôn luôn đúng

áp dụng vào  P ta được như sau

\(\frac{1}{x+x+y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\) luôn đúng :))

\(\frac{1}{x+y+y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}\right)\)

\(\frac{1}{x+y+z+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}\right)\)

Cộng tất cả vào ta được

\(P\le\frac{1}{16}\left(\frac{4}{x}+\frac{4}{y}+\frac{4}{z}\right)\Leftrightarrow P\le\frac{1}{4}\left(x+y+z\right)\)

Thèo đề \(xy+yz+xz=4xyz\Leftrightarrow xy+yz+xz=xyz+xyz+xyz+xyz\)

Tao cũng éo hiểu tại sao nó = nhau được

1 đề sai  , 2 tao sai thế thôi

\(\frac{xy+2x+1}{xy+x+y+1}+\frac{yz+2y+1}{yz+y+z+1}+\frac{zx+2z+1}{zx+z+x+1}\)

Ta có: \(\frac{xy+2x+1}{xy+x+y+1}=\frac{\left(xy+x\right)+\left(x+1\right)}{\left(xy+x\right)+\left(y+1\right)}=\frac{x\left(y+1\right)+\left(x+1\right)}{\left(y+1\right)\left(x+1\right)}=\frac{x}{x+1}+\frac{1}{y+1}\)

Tương tự ta có:

\(\frac{yz+2y+1}{yz+y+z+1}=\frac{y}{y+1}+\frac{1}{z+1}\)

\(\frac{zx+2z+1}{zx+z+x+1}=\frac{z}{z+1}+\frac{1}{x+1}\)

Từ đây ta có biểu thức ban đầu sẽ bằng

\(\frac{x}{x+1}+\frac{1}{y+1}+\frac{y}{y+1}+\frac{1}{z+1}+\frac{z}{z+1}+\frac{1}{x+1}\)

\(\left(\frac{x}{x+1}+\frac{1}{x+1}\right)+\left(\frac{y}{y+1}+\frac{1}{y+1}\right)+\left(\frac{z}{z+1}+\frac{1}{z+1}\right)=1+1+1=3\)

CHÚ Ý: ab+a+b+1=a(b+1)+(b+1)=(a+1)(b+1)

Xét: \(\frac{xy+2x+1}{xy+x+y+1}=\frac{x\left(y+1\right)+x+1}{\left(x+1\right)\left(y+1\right)}=\frac{x}{x+1}+\frac{1}{y+1}\)

Tương tự với 2 biểu thức còn lại ta được:

A=\(\frac{x}{x+1}+\frac{1}{y+1}+\frac{y}{y+1}+\frac{1}{z+1}+\frac{z}{z+1}+\frac{1}{x+1}\)

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