TITLE: Inequality. $abc(a+b+c) > 3abc+ab+bc+ca.$
QUESTION [7 upvotes]: I want to ask you a solution for the following problem. 
Let $a,b,c$ be real numbers, $a,b,c > \frac{1+\sqrt{5}}{2}$. Prove that: 
$$abc(a+b+c) > 3abc+ab+bc+ca.$$
I don't know how "to touch" this problem, I tried to use $AG \geq GM$, but also is a problem because in our inequality appears $>$ and no $\geq $.
thanks:)

REPLY [7 votes]: The inequalities on $a,b,c$ prove that $a^2>a+1,b^2>b+1,c^2>c+1$. Then if you expand the LHS and apply these inequalities you get exactly the desired result.