SOLUTION: Numbers a, b and c are such that a ≥ 0, b ≥ 0 and c ≥ 0. Prove: 2(a^3+b^3+c^3 )≥ab(a+b)+bc(b+c)+ca(c+a)

To prove:


By Cauchy's inequality, the arithmetic mean of any number of 
positive numbers is greater than or equal to the geometric mean.

Let's take two of the cubes to be the same.
The arithmetic mean of ,, and  is greater
than or equal to the geometric mean of those cubes, two of which are the same.






By symmetry we can also prove that







Add all 6 inequalities:



Divide through by 3



Factor out 2 on the left.
Rearrange the terms on the right so we can factor 
pairwise and get the given desired right side: 



On the right, factor the 1st two, middle two and last two terms:

 

Edwin