Prove that

Question

(a+b+c) (ab+bc+ca) ≥ 9abc

 

Answer 1

I think you have to assume that a, b, and c are all ≥ 0 for this to be true.  For example, suppose a = -1, b = -1, c = 1.  Then...

(a + b + c)(ab + bc + ca) = [-1 -1 +1] [(-1)(-1) + (-1)(1) + (1)(-1)] =

(-1)(1 - 1 - 1) = (-1)(-1) = 1

 

But 9abc = 9(-1)(-1)(1) = 9. 

 

So in that case, the statement is not true.  So let's assume that all the numbers are ≥ 0, and that you just left this out of the statement.

 

Let's calculate (a + b + c)(ab + bc + ca) - 9abc.  If that is ≥ 0, then that proves the statement.

 

(a + b + c)(ab + bc + ca) - 9abc =

(a²b + abc + a²c) + (ab² + b²c + abc) + (abc + bc² + ac²) - 9abc =

a²b + a²c + ab² + b²c + bc² + ac² + 3abc - 9abc =

a²b + a²c + ab² + b²c + bc² + ac² - 6abc

 

There are 6 terms that include a square, so lets split up the -6abc among them.  We get:

 

(a²b - abc) + (a²c - abc) + (ab² - abc) + (b²c - abc) + (bc² - abc) + (ac² - abc)

 

Factor each of the terms in parentheses:

 

ab (a-c) + ac (a-b) + ab (b-c) + bc (b-a) + bc (c-a) + ac(c-b)

 

Rearrange:

ab (a-c) + bc (c-a) + ac (a-b) + bc (b-a) + ab (b-c) + ac(c-b) =

ab (a-c) - bc (a-c) + ac (a-b) - bc (a-b) + ab (b-c) - ac(b-c)

 

Factor each group of two:

b (a-c)(a-c) + c (a-b)(a-b) + a (b-c)(b-c) =

b (a-c)² + c (a-b)² + a (b-c)²

 

Now, the square terms are all ≥ 0, and a, b, and c, are all ≥ 0.  That means that the 3 terms in the last expression are all ≥ 0.  So the entire expression is ≥ 0.

 

Going back to the beginning, that means that

 

(a + b + c)(ab + bc + ca) - 9abc ≥ 0

 

So...

 

(a + b + c)(ab + bc + ca) ≥ 9abc

 

And that proves the statement.