SOLUTION: Prove (a+b+c)(ab+ac+bc)>= 9abc where a,b,c are positive. I'm lost as to where to begin this proof

Prove (a+b+c)(ab+ac+bc) ≧ 9abc where a,b,c are positive.

The inequality is symmetric in a,b, and c, so
without loss of generality, we can suppose a≦b≦c

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

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

Substitute those in the left side of the inequality:

[a+(a+p)+(a+p+q)][a(a+p)+a(a+p+q)+(a+p)(a+p+q)]

If you multiply that all the way out and collect terms (whew!),
you will get this:

9a³+18a²p+9a²q+11ap²+11apq+2aq²+2p³+3p²q+pq²

Make the same substitutions in the right side of the inequality:

9a(a+p)(a+p+q)

If you multiply that all the way out and collect terms (not quite
as tedious) you will get this:

9a³+18a²p+9a²q+9ap²+9apq

So the inequality we are to prove becomes

9a³+18a²p+9a²q+11ap²+11apq+2aq²+2p³+3p²q+pq² ≧ 9a³+18a²p+9a²q+9ap²+9apq

For contradiction assume the inequality can be <, that is, assume that
for some non-negative p and q, this holds:

9a³+18a²p+9a²q+11ap²+11apq+2aq²+2p³+3p²q+pq² < 9a³+18a²p+9a²q+9ap²+9apq

Subtracting the right side from the left:

2ap²+2apq+2aq²+2p³+3p²q+pq² < 0

Since all the terms on the left are non-negative,
this is clearly false, so < can never hold, thus 
we have reached a contradiction. ≧ always 
holds and the original inequality is proved true.  QED

Edwin