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

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

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

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

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

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

" = " \(\Leftrightarrow x=y=z=9\)

\(M=\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=\frac{y^3z^3+x^3z^3+x^3y^3}{x^2y^2z^2}=\frac{\left(yz+xz\right)^3+x^3y^3-3xy^2z^3-3x^2yz^3}{x^2y^2z^2}\)

\(=\frac{\left(yz+xz+xy\right)\left[\left(yz+xz\right)^2+xy\left(yz+xz\right)+x^2y^2\right]-3xyz^2\left(xz+yz\right)}{x^2y^2z^2}\)

\(=\frac{0.\left[\left(yz+xz\right)^2+xy\left(yz+xz\right)+x^2y^2\right]-3xyz^2\left(xz+yz\right)}{x^2y^2z^2}\)

\(=\frac{-3xyz^2\left(xz+yz\right)}{x^2y^2z^2}=\frac{-3\left(xz+yz\right)}{xy}=\frac{-3.\left(-xy\right)}{xy}=3\)

Từ đề <=>\(\frac{xyz}{xz+yz}=\frac{xyz}{xy+xz}=\frac{xyz}{xy+zy}\Leftrightarrow xz=xy=zy\)

Có : \(zx=xy\Rightarrow y=z\left(\text{Vì }x\ne0\right),xy=zy\Rightarrow x=z\)

=> x=y=z 

tự tính M :]]

bạn nào t-i-k sai cho tớ làm lại hộ ạ :)

ai t-i-k sai thì làm lại đi :(( 

ta có xy+yz+zx=0=> \(\frac{xy+yz+zx}{xyz}=0\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

đặt \(\frac{1}{x}=a,\frac{1}{y}=b,\frac{1}{z}=c\Rightarrow a+b+c=0\)

ta xét \(a^3+b^3+c^3-3abc=a^3+b^3+3ab\left(a+b\right)+c^3-3ab-3abc\)

           \(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

=> \(a^3+b^3+c^3=3abc\) \(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)

=> \(M=\frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}=\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz.\frac{3}{xyz}=3\)

=> M=3

 Mình nghĩ thế này ạ

xy + 2(yz + xz) =5 => xy + 2yz + 2xz =5

Mình áp dụng bất đẳng thức này nhé :)
Ta có:  \(\left(x-y\right)^2\ge0\forall x,y\)

\(\Rightarrow x^2+y^2\ge2xy\forall x,y\)

\(\Rightarrow\frac{1}{2}\left(x^2+y^2\right)\ge xy\forall x,y\)(1)

Chứng minh tương tự ta được \(y^2+z^2\ge2yz\forall y,z\)(2)

\(x^2+z^2\ge2xz\forall x,z\)(3)

Cộng vế (1) (2) (3) ta được \(\frac{1}{2}\left(x^2+y^2\right)+y^2+z^2+x^2+z^2\ge xy+2yz+2xz\forall x,y,z\)

\(\Rightarrow\frac{1}{2}x^2+\frac{1}{2}y^2+x^2+y^2+z^2+z^2\)\(\ge5\)\(\forall x,y,z\)

\(\Rightarrow\frac{3}{2}x^2+\frac{3}{2}y^2+2z^2\ge5\forall x,y,z\)

nhân cả 2 vế với 2 nè

\(\Rightarrow3x^2+3y^2+4z^2\ge10\forall x,y,z\)

\(\Rightarrow3\left(x^2+y^2\right)+4z^2\ge10\forall x,y,z\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x=y\\y=z;x=z\\xy+2\left(yz+xz\right)=5\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=y=z\\x^2+2.\left(x^2+x^2\right)=5\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=y=z\\5x^2=5\end{cases}\Leftrightarrow\hept{\begin{cases}x=y=z\\x^2=1\end{cases}\Leftrightarrow}}\)x=y=z = 1 hoăc 

Vậy giá trị nhỏ nhất của biểu thức là 10 tại x=y=z=1;-1