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.
 A: Here's another approach. Let $f(x,y,z) = xy+yz+zx-2xyz$, and suppose its maximum value for $x,y,$ and $z$ non-negative reals with sum 1 is $f(a,b,c)$. Because $f$ is symmetric in its arguments, assume without loss of generality that $a \le b \le c$.
A bit of algebra shows that $f(\frac{a+c}{2},b,\frac{a+c}{2}) - f(a,b,c)$ = $\frac{1}{4}(a-c)^2(2a+2c-1)$, which must be less than or equal to zero because $f(a,b,c)$ is a maximum, and $\frac{a+c}{2},b$, and $\frac{a+c}{2}$ are non-negative reals with sum 1. Therefore, either $a=c$ or $2a+2c-1 < 0$. But $1= a+b+c\le a+2c \le 2a+2c$, so $2a+2c-1 \ge 0,$ leaving $a=c$ as the only possibility. This together with the fact that $a\le b \le c$ implies that $a=b=c=\frac{1}{3}$.
A: Here's a calculus solution.
Let's begin with your substitution $x=p+\frac 13, y=q+\frac 13, z=-p-q+\frac 13$.  The problem conditions impose that $p \ge -\frac 13, q \ge -\frac 13, -p-q \ge -\frac 13$.  This is a triangular area shown here: 

The polynomial simplifies to $f(p,q)=2p^2q+2pq^2-\frac{1}{3}(p^2+pq+q^2)+\frac{7}{27}$, and the problem reduces to proving $0\le f(p,q)\le \frac{7}{27}$ over the above region.  We calculate $\frac{\partial}{\partial p}f=\frac{1}{3}(6q-1)(2p+q)$ and $\frac{\partial}{\partial q}f=\frac{1}{3}(6p-1)(2q+p)$.  Setting these each to zero gives four critical points: $(\frac{1}{6},\frac{1}{6}), (\frac{1}{6},\frac{-1}{3}),(\frac{-1}{3},\frac{1}{6}),(0,0)$.  Evaluating $f$ gives $\frac{1}{4}$ at the first three and $\frac{7}{27}$ at the last.
We now must consider the boundary of the triangle.  If $p=-\frac{1}{3}$, $f(p,q)=-q^2+\frac{q}{3}+\frac{2}{9}$, a  downward-facing parabola with maximum at $q=\frac 16$ (already considered) and minimum at the corners of the triangle, namely $(-\frac{1}{3},-\frac{1}{3})$ and $(\frac{2}{3},-\frac{1}{3})$.  At both of these corners $f(p,q)=0$.  By symmetry $q=-\frac{1}{3}$ gives us nothing new.  Lastly we consider $q=\frac{1}{3}-p$.  This gives us $f(p,q)=-p^2+\frac{p}{3}+\frac{2}{9}$, which has maximum at $(\frac{1}{6},\frac{1}{6})$, already considered, and minimum at the corners, already considered.
Hence, among the seven critical points, the unique maximum occurs at $(0,0)$ and the minimum occurs at the three corners.
A: The left inequality:
$$xy+xz+yz-2xyz=(xy+xz+yz)(x+y+z)-2xyz=\sum_{cyc}(x^2y+x^z)+xyz\geq0$$
The right inequality:
$$xy+xz+yz-2xyz\leq\frac{7}{27}\Leftrightarrow(xy+xz+yz)(x+y+z)-2xyz\leq\frac{7(x+y+z)^3}{27}\Leftrightarrow$$
$$\Leftrightarrow\sum_{cyc}(7x^3-6x^2y-6x^2z+5xyz)\geq0\Leftrightarrow6\sum_{cyc}(x^3-x^2y-x^2z+xyz)+\sum_{cyc}(x^3-xyz)\geq0,$$
which is true by Schur and AM-GM.
A: Here's what I think is a much simpler proof than all of the others. 
By the Cauchy Schwarz Inequality, $$(xy+yz+xz)^2\leq (x^2+y^2+z^2)^2\\ \implies xy+yz+xz\leq x^2+y^2+z^2=(x+y+z)^2-2(xy+xz+yz) \\ \implies3(xy+xz+yz)\leq 1 \implies xy+yz+xz\leq \frac{1}{3}$$
By AM-GM,
$$\sqrt[3]{xyz}\leq \frac{x+y+z}{3}\\\leq \frac{1}{3} \\ \implies xyz\leq \frac{1}{27}$$.
The original inequality is equivalent to $$xy+yz+xz\leq \frac{7}{27}+2xyz$$ which is clear given the bounds that we just proved. Equality occurs at $x=y=z=\frac{1}{3}$
A: 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 $-\frac{1}{2}\lbrace a^2(1+2a)+b^2(1+2b)+c^2(1+2c)\rbrace\leq 0$ . 
