f: x+y+z=3

=>x^2+y^2+z^2+2(xy+xz+yz)=9

=>2(xy+yz+xz)=6

=>xy+yz+xz=3

mà x+y+z=3

nên x=y=z=1

e: x^2+y^2+2=2(x+y)

=>(x+y)^2-2xy+2-2(x+y)=0

=>(x+y)(x+y-2)-2(xy-1)=0

=>x=y=1

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)

-có lẽ là x³+y³=x-y-

Vì x,y>0=>x³+y³>0=>x-y>0

Có x²+y²<1<=>(x-y)(x²+y²)<x-y<=>(x-y)(x²+y²)<x³+y³

<=>x³+xy²-x²y-y³<x³+y³<=>x³+y³-x³-xy²+x²y+y³>0

<=>2y³-xy²+x²y>0<=>y(2y²-xy+x²)>0

<=>y[7y²/4+(y/2 - x)²] > 0 (luôn đúng do x,y>0)

\(x^3+y^3+z^3-3xyz=0\)

\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)

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

\(\Leftrightarrow x=y=z\)

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)

Ta có: \(x>y>0\)

\(\Rightarrow x^5-y^5< x^5+y^5\)

\(\Leftrightarrow\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)< x-y\)

\(\Leftrightarrow x^4+x^3y+x^2y^2+xy^3+y^4< 1\)               \(\left(1\right)\)

Lại có: \(x>y>0\)

\(\Rightarrow x^4+y^4< x^4+x^3y+x^2y^2+xy^3+y^4\)               \(\left(2\right)\)

Từ (1) và (2) suy ra \(x^4+y^4< 1\)

Vậy \(x^4+y^4< 1\)

Ta có:  \(x>y>0\)

\(\Rightarrow x^5-y^5< x^5+y^5\)

\(\Leftrightarrow\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)< x-y\)

\(\Leftrightarrow x^4+x^3y+x^2y^2+xy^3+y^4< 1^{\left(1\right)}\)

Lại có: \(x>y>0\)

\(\Rightarrow x^4+y^4< x^4+x^3y+x^2y^2+xy^3+y^4\)(2)

Từ (1) và (2) suy ra : \(x^4+y^4< 1\)

Vậy \(x^4+y^4< 1\)(đpcm)