<s>[INST] For $a$, $b$, $c$, $d$ the sides of a quadrilateral, show $ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0$. (A generalization of IMO 1983 problem 6) 
Let $a$, $b$, $c$, and $d$ be the lengths of the sides of a quadrilateral. Show that
  $$ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0 \tag{$\star$}$$

Background: The well known 1983 IMO Problem 6 is the following:

IMO 1983 #6. Let $a$, $b$ and $c$ be the lengths of the sides of a triangle.
  Prove that $$a^{2}b(a - b) + b^{2}c(b - c) +c^{2}a(c - a)\ge 0. $$

See: here. 
A lot of people have discussed this problem. So this problem (IMO 1983) has a lot of nice methods.
Now I found for quadrilaterals a similar inequality. Are there some nice methods for inequality $(\star)$?
 [/INST] Let $\;$ $LHS=f(a,b,c,d)$. Note that $f(a,b,c,d)>f(a-k,b-k,c-k,d-k)$ for $0<k\le\min\{a,b,c,d\}$. So, WLOG we can take $d=0$ to prove the inequality. In that case we should show $ab^2(b-c)+bc^3\ge0$. If $b\ge c$ obviously we are done. For $c\ge b \ge a$ and $c\ge a \ge b$ cases arranging the inequality shows us $ab^3+bc(c^2-ab)\ge0$. Now, $a\ge c \ge b$ case remained. We will use the inequality $a\ge c$. Let's arrange the inequality again. Then, $ab^2(b-c)+bc^3\ge cb^2(b-c)+bc^3=b^3c-b^2c^2+bc^3\ge0$ $\;$ (by AM-GM $b^3c+bc^3\ge 2b^2c^2$ ) 
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