<s>[INST] If $a, b, c\in\mathbb R^+, $ then prove that $a^3b+b^3c+c^3a\ge abc(a+b+c) .$ While trying to prove it, I proved the following two inequalities:
$a^4+b^4+c^4\ge abc(a+b+c)$ and
$(a^2b+b^2c+c^2a)(ab+bc+ca)\ge abc(a+b+c)^2.$
The later one, on some simplification gives
$a^3b+b^3c+c^3a\ge abc(ab+bc+ca).$
But we can't claim that $ab+bc+ca\ge a+b+c$ for all positive $a, b, c.$ So this doesn't help. So am not quite sure how to approach the inequality in question. Please suggest.. Thanks in advance. (BTW can we use Cauchy-Schwarz's inequality? I tried but couldn't think of a proper choice for two triplets.) 
 [/INST] Suppose $c=\min\{a,b,c\}.$ We have
$$\begin{aligned}a^3b+b^3c+c^3a-abc(a+b+c)&=c(a^3+b^3-a^2b-ab^2)+a(c^3+a^2b-bc^2-ca^2)\\&=c(a+b)(a-b)^2+a(c+a)(a-c)(b-c) \geqslant 0.\end{aligned}$$
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