Inequality involving the side lengths of a quadrilateral If $a$, $b$, $c$ and $d$ are the four sides of a quadrilateral, the problem is to show that $ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0$. I've verified it to be true for quite a large number of values, but can't seem to come up with a proof for it. Does anyone have any ideas, perhaps some inequality results that can be applied to prove it?
 A: Assuming the contrary, we get that the minimal value of 
$$
  f(a,b,c,d)=ab^2(b−c)+bc^2(c−d)+cd^2(d−a)+da^2(a−b)
$$
on the set
$$
  M=\{(a,b,c,d)\colon a,b,c,d\geq 0, a+b+c+d\leq 1, a+b+c\geq d,\dots\}
$$
is negative. By the homogeneity, this value is achieved on $a+b+c+d=1$.
Later we will check that $f$ is nonnegative on the boundary faces $a=0$ and $a+b+c=d$. Thus the minimum is achieved at an interior point of the face defined by $a+b+c+d=1$. By the (Karush--)Kuhn--Tucker theorem, this point satisfies the system
$$
  -\lambda=b^2(b-c)-cd^2+3da^2-2dab=\cdots
$$
for some nonnegative $\lambda$ (the other equalities are obtained by the cyclic permutation of variables). In particular, this means that
$$
  b^3+3da^2\leq b^2c+cd^2+2dab, \;\dots,
$$
which sums up to
$$
  \sum a^3+2\sum ab^2\leq \sum a^2b+2\sum abc
$$
(the sums are also over four cyclic permutations). On the other hand, by AM--GM we have
$$
  a^3+ab^2+bc^2+bc^2\geq 4abc, \quad a^3+ab^2\geq 2a^2b.
$$
Summing up all these we get the converse; thus all these inequalities should come to equalities, and hence $a=b=c=d$; but $f(a,a,a,a)=0$. A contradiction.
It remains to check the inequality on the other boundary faces. If $a=0$ then the inequality looks like $bc^2(d-c)\leq cd^3$; clearly we may assume that $d\geq c$, otherwise the inequality is trivial. Since $b\leq c+d$, we have $ bc^2(d-c)\leq c^2(d^2-c^2)\leq cd\cdot d^2$, as required.
Finally, if $a=b+c+d$ then we may substitute this into our function getting (here Maple was used...) the large sum with many positive terms (including $b^4+bc^3$) and only one negative term $-b^2c^2$. But it is easy to see that $b^4-b^2c^2+bc^3\geq 0$ for all nonnegative $b$ and $c$.
