Re: HSC 2013 4U Marathon
By the product condition, the inequality can be rewritten as:
(abc-1) + (a+b+c) - (ab+bc+ac) > 0
<=> (a-1)(b-1)(c-1) > 0.
Hence either one or all three of the LHS factors must be positive. They cannot all be positive because of the product condition, and so we can conclude that exactly one of a,b,c is greater than one.
Nice solution, I think my one might be yours 'in disguise' since it uses a similar concept
My one was: First conclude that at least 1 of a,b,c needs to be > 1 if it satisfies abc=1. And at least 1 of a,b,c < 1.
This leaves 2 cases, 2 of them > 1, and 1 of them < 1 or
2 of them < 1 and 1 of them > 1, and the second is what we need to prove.
Without loss of generality, let a > 1, b < 1.
(b+c) > (bc-1)(bc+1) )
Since a > 1, then bc < 1, so inequality sign flips when cancelling out
 < (1-b) )
Since b < 1, we can divide
==============
 < a_n < g(n) )
 = \lim_{n \to \infty} f(n) = K )
 )
 \\b\rightarrow \left ( \frac{1}{b} -1\right )\\c\rightarrow \left ( \frac{1}{c}-1 \right ) \\ \\ \therefore \left ( \frac{1}{a}-a \right )\left ( \frac{1}{b} -1\right )\left ( \frac{1}{c}-1 \right ) \leq \left ( \frac{\frac{1}{a}+\frac{1}{b}+ \frac{1}{c}-3}{3} \right )^3 \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\geq \left ( \frac{9-3}{3} \right )^3 ($using part i) \\ \\ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= 8 \\ \\ \therefore LHS_{min} = 8)
