IMO 1984: yz + zx + xy − 2xyz

Notation. Let $q=xy+yz+zx$ and $r=xyz$.

Lower bound. Since $x,y,z\le1$,
$$q-2r=\bigl[xy(1-z)+yz(1-x)+zx(1-y)\bigr]+r\ge0,$$
because $xy(1-z)+yz(1-x)+zx(1-y)=q-3r$.

Upper bound. Expanding, using $x+y+z=1$,
$$(1-2x)(1-2y)(1-2z)=1-2+4q-8r=-1+4q-8r,$$
so $q-2r=\dfrac{1+P}4$ with $P=(1-2x)(1-2y)(1-2z)$. We must show $P\le\frac1{27}$.
- Two of the factors cannot be negative: $1-2x<0$ and $1-2y<0$ would give $x+y>1$, so $z<0$.
- If exactly one factor is negative, the others are non-negative and $P\le0$.
- If none is negative, their sum is $3-2=1$ and AM-GM gives $P\le\left(\frac13\right)^3=\frac1{27}$.

In all cases $P\le\frac1{27}$, hence $q-2r\le\frac{28}{27\cdot4}=\frac7{27}$.

Equality. Upper bound: $x=y=z=\frac13$. Lower bound: e.g. $(1,0,0)$.

Lesson. Rewriting a symmetric expression through $p=x+y+z$, $q$, $r$ (the pqr method) can turn it into a product whose sign analysis and AM-GM finish the job.