# 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 –  B.S. Jan 26 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.

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That was fast. Nice. –  K. Stm. Jan 26 at 19:36
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 at 19:38
Yeah, I was about just to use that, wondering what next, when I saw you already posted the answer. –  K. Stm. Jan 26 at 19:39
@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 at 19:52
sorry for link, new link : math.stackexchange.com/questions/283895/… –  Iuli Jan 26 at 19:58