<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] use the ptolemy inequality !
$ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0$$\Longrightarrow$
$2(ab^3+bc^3+cd^3+da^3)\ge$$b^2\cdot{a}(b+c)+c^2\cdot{b}(c+d)+d^2\cdot{c}(a+d)+a^2\cdot{d}(a+b)$$\Longrightarrow$
$2(ab^3+bc^3+cd^3+da^3)\ge$$b^2\cdot{a}l_{1}+c^2\cdot{b}l_{2}+d^2\cdot{c}l_{1}+a^2\cdot{d}l_{2}$$\Longrightarrow$
$8abcd\ge$$b^2\cdot{a}l_{1}+c^2\cdot{b}l_{2}+d^2\cdot{c}l_{1}+a^2\cdot{d}l_{2}$$\Longrightarrow$
$8\ge$$bl_{1}/cd+cl_{2}/ad+dl_{1}/ab+al_{2}/bc$
if $bl_{1}/cd,cl_{2}/ad,dl_{1}/ab,al_{2}/bc\ge2$
then, $l_{1}l_{2}abcd\ge bc\cdot(cd)^2+ad\cdot(ab)^2+ab\cdot(bc)^2+cd\cdot(ad)^2$
$\Longrightarrow$$ac+bd\ge 4\sqrt{abcd}$$\Longrightarrow$$l_{1}l_{2}\ge ac+bd$
which contradict with the ptolemy inequality !
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