<s>[INST] How To Prove $a^3b＋b^3c＋c^3a>a^2b^2＋b^2c^2＋a^2c^2 $ if $ a > b > c > 0\,$? How to prove this :
$$a^3b＋b^3c＋c^3a>a^2b^2＋b^2c^2＋a^2c^2 $$
if we know: $$ a > b > c > 0 $$
My attempt:
$$\frac {a^3b＋b^3a}{2}>a^2b^2  
   ...(1)$$
$$\frac {b^3c＋c^3b}{2}>c^2b^2...(2)$$
$$\frac {a^3c＋c^3a}{2}>a^2c^2...(3)$$
(1) + (2)+ (3) :
$$\frac {a^3b＋b^3a＋a^3c＋c^3a＋b^3c＋c^3b}{2}>\;a^2b^2＋b^2c^2＋a^2c^2$$
$$a^3b＋b^3c＋c^3a>ab^3＋bc^3＋ca^3\\~\\ab(a^2－b^2)＋bc(b^2－c^2)>ca(a^2-c^2)\\~\\ab(a－b)(a＋b)＋bc(b－c)(b＋c)>ca(a－c)(a＋c)$$
I appreciate your help
 [/INST] We need to prove that:
$$\sum_{cyc}(a^3b-a^2b^2)>0$$ or
$$\sum_{cyc}(2a^3b-2a^2b^2)>0$$ or
$$\sum_{cyc}(a^3b+a^3c-2a^2b^2)+\sum_{cyc}(a^3b-a^3c)>0$$ or
$$\sum_{cyc}ab(a-b)^2+(a-b)(a-c)(b-c)(a+b+c)>0,$$ which is obvious.
I think, BW does not help for the following inequality.

Let $\frac{1}{\sqrt{a}},$ $\frac{1}{\sqrt{b}}$ and $\frac{1}{\sqrt{c}}$ be sides-lengths of a triangle. Prove that:
$$a^3b+b^3c+c^3a\geq a^2b^2+b^2c^2+c^2a^2.$$

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