Áp dụng bđt AM - GM:

\(x^3+1+1\ge3x;y^3+1+1\ge3y;z^3+1+1\ge3z;2x+2y+2z\ge6\sqrt[3]{xyz}=6\).

Cộng vế với vế các bđt trên rồi rút gọn ta có đpcm.

Áp dụng BĐT Cosi:

\(\left(x^3+1+1\right)+\left(y^3+1+1\right)+\left(z^3+1+1\right)\)

\(\ge3\left(x+y+z\right)\)

\(\ge x+y+z+2.3\sqrt[3]{xyz}\)

\(=x+y+z+6\)

\(\Rightarrow x^3+y^3+z^3\ge x+y+z\)

Đẳng thức xảy ra khi \(x=y=z=1\)

Vế trái bằng vế phải nên đẳng thức được chứng minh.

Nếu x ≥ 0, y  ≥  0, z  ≥  0 thì:

x + y + z  ≥  0

x - y 2 + y - z 2 + z - x 2 ≥ 0

Suy ra:

x 3 + y 3 + z 3 - 3 x y z ≥ 0 ⇔ x 3 + y 3 + z 3 ≥ 3 x y z

Hay:  x 3 + y 3 + z 3 3 ≥ x y z

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

=>\(\frac{yz+xz+xy}{xyz}=0\)

=>xy+xz+yz=0

Đặt a=xy; b=xz; c=yz

=>a+b+c=0

=>\(\left(a+b+c\right)^2=0\)

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

=>\(\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ac\right)^2\)

=>\(\left(a^2+b^2+c^2\right)^2=4\left\lbrack a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)\right\rbrack\)

=>\(\left(a^2+b^2+c^2\right)^2=4\left(a^2b^2+b^2c^2+a^2c^2\right)\)

Ta có: \(\left(a^2+b^2+c^2\right)=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)\)

=>\(2\left(a^2b^2+b^2c^2+a^2c^2\right)=\left(a^2+b^2+c^2\right)^2-\left(a^4+b^4+c^4\right)\)

=>\(4\left(a^2b^2+b^2c^2+a^2c^2\right)=2\left(a^2+b^2+c^2\right)^2-2\left(a^4+b^4+c^4\right)\)

DO đó, ta có: \(\left(a^2+b^2+c^2\right)^2=2\left(a^2+b^2+c^2\right)^2-2\left(a^4+b^4+c^4\right)\)

=>\(2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2\)

=>\(2\left(x^4y^4+y^4z^4+x^4z^4\right)=\left(x^2y^2+z^2y^2+x^2z^2\right)^2\)

b:

x+y+z=0

=>x+y=-z

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

\(=\left(x+y\right)^3-3xy\left(x+y\right)-3xzy+z^3\)

\(=\left(-z\right)^3-3xy\cdot\left(-z\right)-3xyz+z^3=z^3+3xyz-3xyz-z^3=0\)