<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] Using “Titu's Lemma” (also called “Engel's form” of the Cauchy-Schwarz inequality) you have for $a, b, c > 0$
$$
\frac{a^3b + b^3c + c^3a}{abc} = \frac{a^2}{c} + \frac{b^2}{a} +\frac{c^2}{b}
\ge \frac{(a+b+c)^2}{c+a+b} = a+b+c
$$
</s>