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

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

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

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

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 a ≥ 0, b  ≥  0, c  ≥  0 thì :

Ta có: \(x^3+y^3+z^3-3xyz\)

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

\(=\left[\left(x+y\right)^3+z^3\right]-\left[3xy\left(x+y\right)+3xyz\right]\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-\left[3xy\left(x+y+z\right)\right]\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-zx-zy+z^2\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-zx-zy+z^2-3xy\right)\)

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

a: =(x+y)^3+z^3-3xy(x+y)-3xyz

=(x+y+z)(x^2+2xy+y^2-xz-yz+z^2)-3xy(x+y+z)

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

b: a+b+c<>0

A=(a+b+c)^3-a^3-b^3-c^3/a+b+c

=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)/(a+b+c)

=a^2+b^2+c^2-ab-ac-bc

=1/2[a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2]

=1/2[(a-b)^2+(b-c)^2+(a-c)^2]>=0

a) 16(12 t 2  +1).

b) Gợi ý x 3   +   y 3 = ( x   +   y ) 3  - 3xy(x + y)

(x + y - z)( x 2   +   y 2   +   z 2  - xy + xz + yz).

x 3  +  y 3  +  z 3  – 3xyz = x + y 3  – 3xy(x + y) +  z 3  – 3xyz

      = [  x + y 3  +  z 3 ] - [ 3xy.(x+ y) + 3xyz]

      = [ x + y 3  +  z 3 ] – 3xy(x + y + z)

      = (x + y + z)[ x + y 2  – (x + y)z +  z 2 ] – 3xy(x + y + z)

      = (x + y + z)( x 2  + 2xy + y 2  – xz – yz + z 2  – 3xy)

      = (x + y + z)( x 2  +  y 2  +  z 2  – xy – xz - yz)

x3 + y3 + z3 - 3xyz

= (x³ + 3x²y + 3xy² + y³) - (3x²y - 3xy²) + z³ - 3xyz

= (x + y)³ - 3xy(x - y) + z³ - 3xyz 

= [(x + y)³ + z³] - 3xy(x + y + z) 

= (x + y + z)³ - 3(x + y)²z - 3(x + y)z² - 3xy(x + y + z)

= (x + y + z)³ - 3z(x + y)(x + y + z) - 3xy(x + y + z) 

= (x + y + z)[(x + y + z)² - 3z(x + y) - 3xy] 

= (x + y + z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy) 

= (x + y + z)(x² + y² + z² - xy - xz - yz)