Biểu thức này không có giá trị cụ thể. Bạn xem lại đề.

Ta có: \(2x^2+xy+2y^2=\dfrac{3}{2}\left(x^2+y^2\right)+\dfrac{1}{2}\left(x^2+2xy+y^2\right)=\dfrac{3}{2}\left(x^2+y^2\right)+\dfrac{1}{2}\left(x+y\right)^2\)

Theo BĐT Bunhacopxky: \(\left(x^2+y^2\right)\left(1+1\right)\ge\left(x+y\right)^2\Rightarrow\dfrac{3}{2}\left(x^2+y^2\right)\ge\dfrac{3}{4}\left(x+y\right)^2\\ \Rightarrow2x^2+xy+2y^2=\dfrac{3}{2}\left(x^2+y^2\right)+\dfrac{1}{2}\left(x+y\right)^2\ge\dfrac{5}{4}\left(x+y\right)^2\\ \Rightarrow\sqrt{2x^2+xy+2y^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)

Chứng minh tương tự:

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

Cộng vế theo vế, ta được: \(P\ge\sqrt{5}\left(x+y+z\right)=\sqrt{5}\cdot1=\sqrt{5}\)

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

Bạn tham khảo nhé

https://hoc24.vn/cau-hoi/cho-cac-so-duong-xyz-thoa-man-xyz1cmrcan2x2xy2y2can2y2yz2z2can2z2zx2x2can5.182722154737

Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)

\(\Leftrightarrow\left(x+y\right)\frac{xy+z\left(x+y+z\right)}{xyz\left(x+y+z\right)}=0\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\)

Vậy x+y=0, y+z=0 hoặc z+x=0

TH1: Nếu x+y=0 => \(x=-y\Rightarrow x^{25}+y^{25}=0\Rightarrow P=0\)

TH2: Nếu y+z=0 => \(y=-z\Rightarrow y^3+z^3=0\Rightarrow P=0\)

TH3: Nếu z+x=0 => \(z=-z\Leftrightarrow z^{2006}-x^{2006}=0\Rightarrow P=0\)

Vậy P=0 

ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2006}\)    (x;y;z khác 0)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)(vì x+y+z=2006)

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

\(\Leftrightarrow\frac{x+y}{xy}=\frac{z-\left(x+y+z\right)}{\left(x+y+z\right).z}\)

\(\Leftrightarrow\frac{x+y}{xy}=\frac{-\left(x+y\right)}{\left(x+y+z\right).z}\)

\(\Leftrightarrow-\left(x+y\right)xy=\left(x+y\right)\left(xz+yz+z^2\right)\)  (vì x;y;z khác 0)

\(\Leftrightarrow\left(x+y\right)\left(xy+yz+xz+z^2\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

=>  x+y=0 hoặc y+z=0 hoặc z+x=0

mà x+y+z=2006 nên

z=2006 hoặc x=2006 hoặc y=2006 

=> đpcm