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.
Claim: (The above.)

Proof: First, notice that 

\begin{eqnarray}
xy + yz + zx - 2xyz &amp;=&amp; xy(1 - z) + yz(1-x) + zx(1-y) + xyz\\
&amp;=&amp; 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 &amp;\ge&amp; \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) $p, q, r &gt; 0$, 

(2) $p, q &gt; 0$ and $r &lt;0$,  

(3) $p &gt; 0$ and $q, r &lt;0$, or 

(4) $p, q, r &lt; 0$.

(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 $0 &lt; |p|, |q| &lt; |r|$. But then $|p|^3 + |q|^3 &lt; |r|^3$, 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 $p, q &lt; p + q \le \frac{1}{3}$,  

\begin{eqnarray} 5(p^2 + q^2 + (p+q)^2) + 8(p^3 + q^3 - (p+q)^3) &amp;=&amp; 10(p^2 + q^2 + pq) - 24(p^2q + q^2p)\\
 &amp;=&amp; 10pq + p^2(10 - 24q) + q^2(10 -24p)\\ &amp;\ge&amp; 10pq + p^2(10 -24(\frac{1}{3})) + q^2(10 -24(\frac{1}{3}) \\ &amp;=&amp; 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.
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 $x=y=z=\frac{1}{3}$