Triangle Problem: Prove (a+1)(b+1)(c+1) < 4

[FeDeX_LaTeX]

Homework Statement

Let a, b and c be the lengths of the sides of a triangle. Suppose that ab + bc + ca = 1. Show that (a+1)(b+1)(c+1) < 4.

Homework Equations

N/A (Olympiad problem)

The Attempt at a Solution

I've made a few inroads at this problem, but whether or not they're going to take me in the right direction, I'm not sure. I used the arithmetic/harmonic mean inequality to show that:

$$9abc \leq a + b + c$$

and I've used substitutions from their original equation to arrive at

$$8(a+b+c) \leq (a+b)(1+9c^2) + (b+c)(1+9a^2) + (a+c)(1+9b^2)$$

but I'm puzzled at where to go from here. I don't think it would be useful to consider specific cases such as the equilateral triangle (a = b = c), since we appear to be trying to prove a general case, and I know I mustn't start from their solution and show that it's equal to the original equation.

Am I taking the right approach? Is a geometric approach better? Any hints/tips anyone can give, without giving me the answer?

Thanks.

 

[FeDeX_LaTeX]

I recognise that

$$(a + 1)(b + 1)(c + 1) - 2(ab + bc + ca) - 2 = (a - 1)(b - 1)(c - 1)$$

and so

$$(a + 1)(b + 1)(c + 1) - 4(ab + bc + ca) = (a - 1)(b - 1)(c - 1)$$

...I feel like I'm close, but not sure where to place the final step.

EDIT: Ugh, it's obvious... clearly the RHS is smaller than zero, so...

$$(a+1)(b+1)(c+1) < 4(ab + bc + ca)$$
$$(a+1)(b+1)(c+1) < 4$$

Done.

 

[FeDeX_LaTeX]

How do I show that a, b and c are all smaller than 1 though? It's a requirement for the above. I thought it could just be assumed but it is not as simple as that.