IMO 1984: Prove that $0 ≤ yz + zx +xy −2xyz ≤ \frac {7}{27}$, where $x,y$ and $z$ are non-negative real numbers for which $x + y + z = 1.$

I checked your solution. Your solution is right.

I like the following way.

The homogenization helps.

By AM-GM $$xy+xz+yz-2xyz=(x+y+z)(xy+xz+yz)-2xyz\geq9xyz-2xyz=7xyz\geq0.$$ Also, $$xy+xz+yz-2xyz\leq\frac{7}{27}$$ it's $$(xy+xz+yz)(x+y+z)-2xyz\leq\frac{7}{27}(x+y+z)^3$$ or $$\sum_{cyc}(7x^3-6x^2y-6x^2z+5xyz)\geq0,$$ which is true by Schur and AM-GM.

Assume $x= \max\{\,x,\,y,\,z\,\}\,\therefore\,3\,x\geqq 1$.

For $z= 1- x- y$, we need to prove

$$x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}\geqq 0$$

We have

$$\because\,x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}=$$

$$= \left \{ 4(\,1- 2\,x\,)y^{\,2}- 4(\,x- 1\,)(\,2\,x- 1\,)y+ \frac{1}{27}(\,5- 9\,x\,)^{\,2} \right \}+$$

$$+ \frac{1}{3}(\,2\,x- 1\,)(\,3\,y- 1\,)(\,3\,x+ 3\,y- 2\,)$$

$$\because\,x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}=$$

$$= \frac{1}{27}(\,3\,x- 1\,)^{\,2}- \frac{1}{9}(\,2\,x- 1\,)(\,3\,y- 1\,)(\,3\,x+ 3\,y- 2\,)$$

$$\because\,x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}=$$

$$= \frac{1}{108}(\,1- 3\,x\,)^{\,2}(\,6\,x+ 1\,)+ \frac{1}{4}(\,1- 2\,x\,)(\,x+ 2\,y- 1\,)^{\,2}$$

Because

$$4(\,1- 2\,x\,)y^{\,2}- 4(\,x- 1\,)(\,2\,x- 1\,)y+ \frac{1}{27}(\,5- 9\,x\,)^{\,2}\geqq 0$$

$$\because\,{\rm discriminant}[\,4(\,1- 2\,x\,)y^{\,2}- 4(\,x- 1\,)(\,2\,x- 1\,)y+ \frac{1}{27}(\,5- 9\,x\,)^{\,2},\,y\,]=$$

$$= \frac{32}{27}(\,2\,x- 1\,)(\,3\,x- 1\,)^{\,3}\leqq 0$$

So

$$x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}\geqq 0$$

$\lceil$ ANOTHER!way $\rfloor$

As @Elvin's above, we can choose $1- 2\,x\geqq 0$ (or $\because\,(\,1- 2\,x\,)(\,1- 2\,y)\leqq 0\,\because\,x+ y< 1$).

$$\because\,x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}=$$

$$= \frac{1}{108}(\,1- 3\,x\,)^{\,2}(\,6\,x+ 1\,)+ \frac{1}{4}(\,1- 2\,x\,)(\,x+ 2\,y- 1\,)^{\,2}\geqq 0$$