Ta có : \(\left(x+y+z\right)^2=x^2+y^2+z^2\)

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

        \(\Rightarrow2\left(xy+yz+zx\right)=0\)

        \(\Rightarrow xy+yz+zx=0\)

        \(\Rightarrow\frac{xy}{xyz}+\frac{yz}{xyz}+\frac{zx}{xyz}=0\)( Chia 2 vế cho xyz )

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

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

Ta lại có : \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\left(\frac{1}{x}+\frac{1}{y}\right)^3-\left(\frac{3}{x^2y}+\frac{3}{xy^2}\right)+\frac{1}{z^3}\)

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

                \(=-\frac{3}{xy}\cdot-\frac{1}{z}\)\(=\frac{3}{xyz}\)

                 \(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)         ( đpcm )

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

\(\Leftrightarrow xy+yz+zx=0\)

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

Ta lại co:

\(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}-\frac{3}{xyz}=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-\frac{1}{xy}-\frac{1}{yz}-\frac{1}{zx}\right)=0\)

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

Ta có:

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

\(\Leftrightarrow xy+yz+zx=0\)

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

Ta lại có:

\(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}-\frac{3}{xyz}=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-\frac{1}{xy}-\frac{1}{yz}-\frac{1}{zx}\right)=0\)

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

P = x^3 (z-y^2) +y^3(x-z^2)+z^3(y-x^2)+xyz(xyz-1) 
= -x^3 (y^2-z) +y^3x-y^3z^2 +z^3y-z^3x^2+x^2y^2z^2-xyz 
= -x^3 (y^2-z)+(y^3x-xyz)-(y^3z^2-z^3y)+(x^2y^2... 
= -x^3 (y^2-z)+xy(y^2-z)-yz^2(y^2-z)+x^2z^2(y^2... 
= (y^2-z)(-x^3+xy-yz^2+x^2z^2) 
= (y^2-z)[-x(x^2-y)+z^2(x^2-y)] 
= (y^2-z)(x^2-y)(z^2-x) = b. a. c ko phụ thuộc vào biến

Đề sai nhé, \(\dfrac{z^2}{x+1}\) mới đúng nha

\(\dfrac{x^2}{y+1}+\dfrac{y^2}{z+1}+\dfrac{z^2}{x+1}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+3}\left(\text{Svácxơ}\right)\)

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

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)

Ta có: \(x+y+z\ge3\sqrt[3]{xyz}=3\)

\(\Rightarrow x+y+z+3\le2\left(x+y+z\right)\)

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{x^2}{1+y}+\frac{y^2}{1+z}+\frac{z^2}{1+x}\geq \frac{(x+y+z)^2}{1+y+1+z+1+x}=\frac{(x+y+z)^2}{(x+y+z)+3}\)

Áp dụng BĐT Cauchy:

\(x+y+z\geq 3\sqrt[3]{xyz}=3\)

Do đó:

\(\frac{x^2}{1+y}+\frac{y^2}{1+z}+\frac{z^2}{1+x}\geq \frac{(x+y+z)^2}{(x+y+z)+3}\geq \frac{(x+y+z)^2}{(x+y+z)+(x+y+z)}=\frac{x+y+z}{2}\geq \frac{3}{2}\)

Ta có đpcm.

Dấu "=" xảy ra khi $x=y=z=1$

P/s: Bạn chú ý lần sau gõ tiêu đề bằng công thức toán !!!

\(\dfrac{\sqrt{1+x^3+y^3}}{xy}>=\sqrt{\dfrac{3}{xy}}\)

\(\dfrac{\sqrt{1+y^3+z^3}}{yz}>=\sqrt{\dfrac{3}{yz}}\)

\(\dfrac{\sqrt{1+z^3+x^3}}{xz}>=\sqrt{\dfrac{3}{xz}}\)

=>\(VT>=\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)=3\sqrt{3}\)