# Inequality. $abc(a+b+c) > 3abc+ab+bc+ca.$

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:)

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I really like the inequalities you have shared here on the site. May I ask you what is your source? Is there a certain book or note for that? Thanks +1 –  Babak S. Jan 26 '13 at 19:59

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.
The whole idea was to find out how can we use the fact that all the variables are greater than $(1+\sqrt{5})/2$. –  Beni Bogosel Jan 26 '13 at 19:38
@BeniBogosel Thanks :) I tried to use $AM\geq GM$ but nothing. I like this problem, I thought it must be a nice idea, but I didn't have it. If you have time, can you give me please, an idea for this inequality: math.stackexchange.com/users/33954/iuli . Thanks, best wishes ! –  Iuli Jan 26 '13 at 19:52