SOLUTION: show a(b^2)+ b(c^2) + c(a^2) >= 3abc I'm lost as to how to begin proof

ab² + bc² + ca² ≧ 3abc

Note:  the inequality does not hold if we allow negative 
numbers because if a=-1, b=5, and c=1,

-1·5² + 6·1² + 1·(-1)² ≧ 3·(-1)(1)
  -25 + 6 + 1 ≧ -3
          -18 ≧ -3

That is false.  So you must mean that a,b, and c are 
non-negative.

So I will assume that a,b,c are all non-negative.
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The inequality is symmetric in a,b, and c.  Therefore
without loss of generality we can assume a≦b≦c.

Then there exist non-negative numbers p and q such that

b = a+p, and c = a+p+q

Substituting those for b and c in the left side of the 
inequality gives:

a(a+p)² + (a+p)(a+p+q)² + (a+p+q)a²

Multiplying that out and collecting like terms gives:

3a³+6a²p+3a²q+4ap²+4apq+aq²+aq²+p³+2p²q+pq² 

Making the same substitutions in the right side of the 
inequality gives:

3a(a+p)(a+p+q)

Multiplying that out and collecting like terms gives:

3a³+6a²p+3a²q+3ap²+3apq

So the inequality which we are to prove becomes:

3a³+6a²p+3a²q+4ap²+4apq+aq²+aq²+p³+2p²q+pq² ≧ 3a³+6a²p+3a²q+3ap²+3apq

For contradiction, assume that 

3a³+6a²p+3a²q+4ap²+4apq+aq²+aq²+p³+2p²q+pq² < 3a³+6a²p+3a²q+3ap²+3apq

holds for some non-negative a, p, and q

Subtracting the right side from the left side we get

ap²+apq+2aq²+p³+2p²q+pq² < 0

But this can never hold because the left side is non-negative, 
so we have reached a contradiction.  Therefore the original 
inequality holds.

Edwin