Verifying a proof that if $x,y,z \geq 0$ and $x+y+z = 1$, then $0 \le xy + yz + zx - 2xyz \le \frac{7}{27}$ I was working some recreational problems from a book (The Art and Craft Of Problem Solving, Zeitz) and came across one from the '84 IMO:

Suppose that $x, y, z$ are non-negative reals, with $x + y + z = 1$. Prove that 
  $$0 \le xy + yz + zx - 2xyz \le \frac{7}{27}.$$

I'm afraid that I'm here today to ask for a proof check. I ask because I was able to prove the claim without using any sophisticated inequalities, and - although I have checked it myself - I can't help but feel a bit suspicious. 
I have posted my solution as an answer below. Of course, I would love it if somebody would give it a careful read.
Shorter, more elegant solutions are also welcome.

For proving the second inequality, take $ x=a+\frac{1}{3}, y=b+\frac{1}{3}, z=c+\frac{1}{3}$. Since $x+y+z=1$, so $a+b+c=0$. Since $x,y,z\geq 0$, so $a,b,c\geq -\frac{1}{3}$.By simple algebraic manipulation we get,
$$xy+yz+zx-2xyz=\frac{2}{3}(ab+bc+ca-3abc)+\frac{7}{27}$$
We just need to show that $ab+bc+ca-3abc\leq 0$. Since $a+b+c=0$, so $a^2+b^2+c^2=-2(ab+bc+ca)$ and $a^3+b^3+c^3=3abc$. Thus,
$$ab+bc+ca-3abc=-\frac{1}{2}(a^2+b^2+c^2)-(a^3+b^3+c^3)$$
which is equal to $-\frac{1}{2}\lbrace a^2(1+2a)+b^2(1+2b)+c^2(1+2c)\rbrace\leq 0$ . 
