Claim: (The above.)
Proof: First, notice that
\begin{eqnarray}
xy + yz + zx - 2xyz &=& xy(1 - z) + yz(1-x) + zx(1-y) + xyz\\
&=& xy(x+y) + yz(y+z) + zx(z+x) + xyz
\end{eqnarray}
since $x + y + z = 1$; since we can re-write our original expression as a sum of all positive terms, we plainly have
$$0 \le xy + yz + zx - 2xyz$$
which was the first part of the claim. For the second part, we first notice that
$$\frac{1}{3}(x + y + z)^3 = \frac{1}{3}(x^3 + y^3 + z^3) + (x + y + z)(xy + yz + yz) -xyz;$$
since $x + y + z = 1$, we get
\begin{eqnarray}
\frac{1}{3} = \frac{1}{3}(x^3 + y^3 + z^3) + [(xy + yz + yz) -2xyz] + xyz,
\end{eqnarray}
so we can re-write our original inequality as
$$\frac{1}{3} - \frac{1}{3}(x^3 + y^3 + z^3) - xyz \le \frac{7}{27}$$
or, just as well,
$$ \frac{1}{3}(x^3 + y^3 + z^3) + xyz \ge \frac{2}{27}.$$
Now, unfortunately, comes a wave of algebra, which I will not do out here. (I will just show you the results, which I checked carefully several times.)
We let $x = \frac{1}{3} + p, y = \frac{1}{3} + q, z = \frac{1}{3} + r$. Importantly, we have $p + q + r = 0$.
After a wave of algebra, our inequality can be re-written as
\begin{eqnarray} \frac{2}{27} + (p^3 + q^3 + r^3) + (p^2 + q^2 + r^2) + \frac{1}{3}(pq + pr + qr) + pqr &\ge& \frac{2}{27} \\ \iff (p^3 + q^3 + r^3) + (p^2 + q^2 + r^2) + \frac{1}{3}(pq + pr + qr) + pqr \ge 0. \end{eqnarray}
Recall that I earlier pointed out (with $x, y,$ and $z$ as the variables) that
$$\frac{1}{3}(p + q + r)^3 = \frac{1}{3}(p^3 + q^3 + r^3) + (p + q + r)(pq + pr + qr) -pqr;$$
we may thus conclude that $pqr = \frac{1}{3}(p^3 + q^3 + r^3)$. Similarly, expanding $(p+q+r)^2$ yields $pq + pr + qr = -\frac{1}{2}(p^2 + q^2 + r^2)$. Substituting, we see that
\begin{eqnarray} (p^3 + q^3 + r^3) + (p^2 + q^2 + r^2) + \frac{1}{3}(pq + pr + qr) + pqr \ge 0 \\ \iff \frac{4}{3}(p^3 + q^3 + r^3) + \frac{5}{6}(p^2 + q^2 + r^2) \ge 0 \end{eqnarray}
Finally, we obtain the equivalent inequality we will make a stand with:
$$ 5(p^2 + q^2 + r^2) + 8(p^3 + q^3 + r^3) \ge 0. \qquad(*) $$
Now, we're almost done. We need only consider the signs of $p, q,$ and $r$. WLOG, we must have
(1) $p, q, r > 0$,
(2) $p, q > 0$ and $r <0$,
(3) $p > 0$ and $q, r <0$, or
(4) $p, q, r < 0$.
(We ignore the cases where any of $p, q,$ or $r$ are zero because - since $p + q + r = 0$ - the inequality $(*)$ then becomes trivial.) Clearly, (1) and (4) are impossible (since $p + q + r$ must sum to something non-zero in those cases). Let us consider case (3): if $p$ and $q$ are both negative, then, in order that $p + q + r = 0$, we must have $0 < |p|, |q| < |r|$. But then $|p|^3 + |q|^3 < |r|^3$, and so $(p^3 + q^3 + r^3)$ is positive; the inequality $(*)$ then plainly holds.
So, the only case to worry about is (2). In that case, noting of course that $p + q = -r$, and also that $-\frac{1}{3} \le r$ (since $x,y $ and $z$ were positive) and therefore $p, q < p + q \le \frac{1}{3}$,
\begin{eqnarray} 5(p^2 + q^2 + (p+q)^2) + 8(p^3 + q^3 - (p+q)^3) &=& 10(p^2 + q^2 + pq) - 24(p^2q + q^2p)\\
&=& 10pq + p^2(10 - 24q) + q^2(10 -24p)\\ &\ge& 10pq + p^2(10 -24(\frac{1}{3})) + q^2(10 -24(\frac{1}{3}) \\ &=& 10pq + 2p^2 + 2q^2 \ge 0. \end{eqnarray}
Thus, the inequality $(*)$ is proven in all cases; and so is the inequality $xy + yz + zx - 2xyz \le \frac{7}{27}$. The claim follows.