The question
Let the real numbers $a,b,c$ be such that $abc(a+b+c)=1$
a) Give an example of a triplet $(a,b,c)$ that verifies the given relation.
b) If, in addition, $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}=a^3b+b^3c+c^3a+2$ and $a+b+c$ is different from zero show that $a=b=c$
The idea
a) I managed to do point a finding the triplet where $a=b=c$
$a=b=c => 3a^4=1 =>a=b=c=\sqrt{\sqrt{\frac{1}{3}}}$
b) For point b the first thing that came into my ming is to show that $ \frac{a}{c}+\frac{b}{a}+\frac{c}{b} \leq a^3b+b^3c+c^3a+2$ by the inequality means and this would make the inequality only when $a=b=c$ but I couldn't
$\frac{a}{c}+\frac{b}{a}+\frac{c}{b}=a^3b+b^3c+c^3a+2=a^3b+b^3c+c^3a+2abc(a+b+c)$
and processing this equality I got to
$\frac{a^2b+b^2c+c^2a}{abc}=abc(\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}+2a+2b+2c) => \frac{1}{b^2c}+\frac{1}{a^2c}+\frac{1}{b^2a}=\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}+2a+2b+2c$
I don't know what to do forward! Hope one of you can help me! Thank you!
Rewrite $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}=a^3b+b^3c+c^3a+2$ as
$$(\frac{a}{c}+\frac{b}{a}+\frac{c}{b})abc(a+b+c)=a^3b+b^3c+c^3a+2abc(a+b+c).$$
It is equivalent to:
$$\sum a^3b+ab^2c+a^2c^2=\sum a^3b+2a^2bc$$
$$\sum a^2c^2=\sum a^2bc$$
Due to Muirhead equality happens when and only when $a=b=c$. All sums are cyclic.
Edit
You could do this to avoid referencing Muirhead. Add these three AM-GMs and divide by $2$: $$a^2b^2+a^2c^2\ge 2a^2bc$$ $$b^2c^2+b^2a^2\ge 2ab^2c$$ $$c^2a^2+c^2b^2\ge 2abc^2$$