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 $(\star)$?
 A: 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}
$
A: WLOG, assume that $a = \max(a, b, c, d)$.
Let $x = a- b, \ y = a-c, \ z = a-d$. Then $x, y, z \ge 0$.
Let  $w = b+c+d - a$. Then $w > 0$ since $a, b, c, d$ are the sides of a quadrilateral.
The inequality is written as $$\frac{1}{4}Aw^2 + \frac{1}{4}Bw + \frac{1}{4}C \ge 0$$ where
\begin{align}
A &= 2\, x^2 - x\, y - 2\, x\, z + 2\, y^2 - y\, z + 2\, z^2,\\
B &= 2\, x^3 + 4\, x^2\, y - 2\, x\, y^2 - 4\, x\, y\, z + 2\, y^3 + 4\, y^2\, z - 2\, y\, z^2 + 2\, z^3,\\
C &= (x+3z)y^3 - 4xzy^2 + (3x^3+3x^2z-3xz^2+z^3)y.
\end{align}
It suffices to prove that $A, B, C \ge 0$.
We have
\begin{align}
A = (x^2 - xy + y^2) + (y^2 - yz + z^2) + (z^2 - 2zx + x^2) \ge 0
\end{align}
and
\begin{align}
B &= 2x^3 + 3x^2y + (x^2y - 2xy^2 + y^3) - 2(2xz)y + y^3 + 3y^2z + z^3 + (y^2z - 2yz^2 + z^3)\nonumber\\
&\ge 2x^3 + 3x^2y - 2(x^2+z^2)y + y^3 + 3y^2z + z^3\nonumber\\
&= 2x^3 + x^2y  + y^3 + 2y^2z +  (y^2z + z^3 - 2yz^2)\nonumber\\
&\ge 0.
\end{align}
And $C\ge 0$ follows from
\begin{align}
&4(x+3z)(3x^3+3x^2z-3xz^2+z^3) - (4zx)^2\nonumber\\
=\ & 12x^4+48x^3z+8x^2z^2-32xz^3+12z^4\nonumber\\
=\ & (12x^4 - 16x^3z + 24x^2z^2) + 64x^3z - 16x^2z^2 - 32xz^3 + 12z^4\\
\ge \ & 64x^3z - 16x^2z^2 - 32xz^3 + 12z^4 \nonumber \\
= \ & 4z(4x+3z)(2x-z)^2\nonumber\\
\ge \ &0.
\end{align}
We are done.  
A: 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 !
A: 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 $b^3c+bc^3\ge 2b^2c^2$ ) 
