<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] WLOG $a = \max(a,b,c,d)$
Let $t = \frac{1}{2} (b+c+d-a) \ge 0$ because $a$,$b$,$c$,$d$ are sides of a quadrilateral
Then decreasing $(a,b,c,d)$ simultaneously by $t$ reduces the desired expression
And $a-t = (b-t)+(c-t)+(d-t)$
Thus it suffices to minimize the expression when $a=b+c+d$, which reduces to:
$
\begin{align}
&(b+c+d) b^2 (b-c) + b c^2 (c-d) - c d^2 (b+c) + d (b+c+d)^2 (c+d) \\
& = \left( b^4 - b^2 c^2 + b^3 d - b^2 c d \right) + \left( b c^3 - b c^2 d \right) - c d^2 (b+c) + d (b+c+d) (b+c+d) (c+d) \\
& \ge \left( b^4 - b^2 c^2 + b^3 d - b^2 c d \right) + \left( b c^3 - b c^2 d \right) - c d^2 (b+c) + \Big( d (b+c) d c + d (b+c) (b+c) c \Big) \\
& \ge \left( b^4 - b^2 c^2 + b^3 d - b^2 c d \right) + \left( b c^3 - b c^2 d \right) + b (b+c) c d \\
& \ge b^4 - b^2 c^2 + b c^3 \\
& \ge 0 \quad\text{because} \quad \frac{1}{3} ( b^4 + 2 b c^3 ) \ge b^2 c^2 \quad \text{by AM-GM}
\end{align}
$
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