How To Prove $a^3b＋b^3c＋c^3a>a^2b^2＋b^2c^2＋a^2c^2 $ if $ a > b > c > 0\,$?

Question

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

Answer 1

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: $$a^3b+b^3c+c^3a\geq a^2b^2+b^2c^2+c^2a^2.$$

Answer 2

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

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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 $>0$.

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Here are two exercises.

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

2. 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)