x²+y²+z²=1 => x;y;z \(\le1\)(1)

1= (x+y+z)²= x²+y²+z²+ 2(xy+yz+zx) = 1+ 2(xy+yz+zx) => xy+yz+zx=0 => xy= z(-y-x) = z(z-1)

x³+y³ =1 <=> (x+y)(x²+y² -xy)=1 <=> (1-z)(1-z²-z(z-1))=1 <=> (z-1)(2z2-z-1)= 2z³ -3z² =0 <=> z=0 hoặc z= \(\frac{3}{2}\)(loại vì lớn hơn 1)

 z=0 => x+y=1; xy= 0;

y=y(x+y) = xy+ y² = y²

=> x+y² +z³ = x+ y+ 0 = 1 (điều phải chứng minh)

\(x^2+y^2+z^2+3+2\left(x+y+z\right)=0\)
\(\Leftrightarrow x^2+y^2+Z^2=2x+2y+2z=0\) 

\(\Leftrightarrow x^2-2x+1+y^2-2y+1+z^2-2z+1=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-1\right)^2=0\\\left(z-1\right)^2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x-1=0 \\y-1=0\\z-1=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x-1=0\\y-1=0\\z-1=0\end{cases}}\)

Vậy \(x=y=z=1\)

Với mọi x;y;z ta luôn có:

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2\ge2xy+2yz+2zx\)

\(\Leftrightarrow3x^2+3y^2+3z^2\ge x^2+y^2+z^2+2xy+2yz+2zx\)

\(\Leftrightarrow3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)

\(\Leftrightarrow x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2=\frac{1}{3}\) (đpcm)

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

Từ x+y+z=1 => 1-x = y+z

Áp dụng BĐT \(\left(a+b\right)^2\ge4ab\), ta có :  \(4\left(1-x\right)\left(1-y\right)\left(1-z\right)=4\left(y+z\right)\left(1-z\right)\left(1-y\right)\le\left[\left(y+z\right)+\left(1-z\right)\right]^2.\left(1-y\right)\)

\(\Rightarrow4\left(y+z\right)\left(1-y\right)\left(1-z\right)\le\left(1+y\right)^2\left(1-y\right)=\left(1+y\right)\left(1-y^2\right)\le1+y\)

\(\Rightarrow1+y=x+2y+z\ge4\left(1-x\right)\left(1-y\right)\left(1-z\right)\)(ĐPCM)