Schur's inequality and a bound with a + b + c = 1

1. The expression is symmetric, so assume $a\ge b\ge c$. Then
$$a(a-b)(a-c)+b(b-a)(b-c)=(a-b)^2(a+b-c)\ge0,$$
and $c(c-a)(c-b)=c(a-c)(b-c)\ge0$. Hence the sum is $\ge0$. Equality iff $a=b=c$, or $a=b$ and $c=0$.

2. Expanding, Schur's inequality reads $a^3+b^3+c^3+3abc\ge\sum_{\rm sym}a^2b$. Meanwhile
$$(a+b+c)^3=\sum a^3+3\sum_{\rm sym}a^2b+6abc,\qquad(a+b+c)(ab+bc+ca)=\sum_{\rm sym}a^2b+3abc.$$
So
$$(a+b+c)^3+9abc-4(a+b+c)(ab+bc+ca)=\sum a^3-\sum_{\rm sym}a^2b+3abc\ge0.$$

3. Upper bound. With $p=1$, part 2 gives $q\le\frac{1+9r}4$. By AM-GM $r\le\frac1{27}$, hence
$$q-2r\le\frac{1+9r}4-2r\le\frac14+\frac r4\le\frac7{27}.$$
Equality at $x=y=z=\frac13$.

Lower bound. Let $z\le\frac13$ be the smallest. Then
$$xy+yz+zx-2xyz=xy(1-2z)+z(x+y)\ge0.$$

Remark. Part 3 is IMO 1984 problem 1.

Lesson. When a symmetric degree-3 inequality has equality at $a=b=c$ *and* at $(1,1,0)$, think Schur.