Question

Prove the identity x^3 + y^3 + z^3 - 2xyz = ( x+y+z) (x^2 + y^2 + z^2 - xy - yz -XX)

Answer 1

Using (x−y)2≥0(x−y)2≥0 one gets x2+y2≥2xyx2+y2≥2xy. Since each xx, yy, and zz are positive, the following three equations can be obtained

x3+xy2≥2x2yx3+xy2≥2x2yx2y+y3≥2xy2x2y+y3≥2xy2x2z+y2z≥2xyzx2z+y2z≥2xyz

Add these three gives the following:

A. x3+y3+x2z+y2z≥x2y+xy2+2xyzx3+y3+x2z+y2z≥x2y+xy2+2xyz

Similarly starting from (y−z)2≥0(y−z)2≥0 the following three equations can be obtained

y3+yz2≥2y2zy3+yz2≥2y2zy2z+z3≥2yz2y2z+z3≥2yz2y2x+z2x≥2xyzy2x+z2x≥2xyz

Add these three gives the following:

B. y3+z3+y2x+z2x≥y2z+yz2+2xyzy3+z3+y2x+z2x≥y2z+yz2+2xyz

Similarly starting from (z−x)2≥0(z−x)2≥0 the following three equations can be obtained

z3+zx2≥2z2xz3+zx2≥2z2xz2x+x3≥2zx2z2x+x3≥2zx2z2y+x2y≥2xyzz2y+x2y≥2xyz

Add these three gives the following:

B. z3+x3+z2y+x2y≥z2x+zx2+2xyzz3+x3+z2y+x2y≥z2x+zx2+2xyz

Adding adding equations A, B and C gives

2(x3+z3+x3)≥6xyz2(x3+z3+x3)≥6xyz

or

x3+z3+x3≥3xyz