<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] Here is a no-brainer approach suggested by Calvin Lin, compared to Michaels Rozenberg's magical answer.  

We will use a positive variable to stand for the difference between two given variables and then eliminate the larger one of the two given variables.
For example since $b>c$, we let $b=c+x$ where $x>0$. We will substitute $c+x$ for $b$, eliminating $b$. We introduce a new "free" variable, $x$, which is very easy to use since the only condition on $x$ is $x>0$.
Similarly, let $a=b+y$ where $y>0$. So, we have $a=c+x+y$. We will also  substitute $c+x+y$ for $a$, eliminating $a$ and welcoming the "free" variable $y$.
With some straightforward labor, we can find $\text{LHS}-\text{RHS}$ is
$$\begin{aligned}
&\quad a^2b(a-b) + b^2c(b-c) + c^2a(c-a)\\
&=(c+x+y)^2(c+x)y + (c+x)^2cx + c^2(c+x+y)(-x-y)\\
&=c^2 x^2 + c^2 x y + c^2 y^2 + c x^3 + 3 c x^2 y + 4 c x y^2 + c y^3 + x^3 y + 2 x^2 y^2 + x y^3
\end{aligned}$$
which is obvious $>0$ since all variables are $>0$.

Here are two exercises.

*

*Suppose $a>b>c>0$. Prove $a^3b+b^3c+c^3a > ab^3+bc^3+ca^3.$


*Suppose $a>b>c>d>0$. Prove $$a^3b+b^3c+c^3d+d^3a > a^2b^2+b^2c^2+c^2d^2+d^2a^2$$
(The computation might be long and boring. The point is no magic trick is needed)
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