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.

  • $\begingroup$ @TMM I fear that your more descriptive and legitimate title may not rouse the same interest that my goofy one did. (Thanks, though.) Also... did you look the proof over??? $\endgroup$
    – Chris
    Commented Apr 27, 2013 at 23:54
  • $\begingroup$ True. It is better to balance marketing and objectiveness in titles. Also, we want them to be informative. $\endgroup$
    – Pedro
    Commented Apr 28, 2013 at 0:05
  • $\begingroup$ There's always a trade-off between rousing interest and being informative, but generally uninformative titles are frowned upon and edited whenever possible. Something like "plz help me with a hard question!!1!!1!" is also not a proper title. You should also keep in mind that people who have the same question later on should be able to find the question with a site search. $\endgroup$
    – TMM
    Commented Apr 28, 2013 at 12:58
  • $\begingroup$ (And of course it's not hard to draw attention to your question with uninformative titles, but if we allow you to do that, then we should also allow others to do that... in which case the frontpage will soon be a mess with titles like "HELP ME PLEASE NOW!" and "This is the HARDEST QUESTION EVAR".) $\endgroup$
    – TMM
    Commented Apr 28, 2013 at 13:02

6 Answers 6


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.


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 $a=b=c=\frac{1}{3}$.


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:

region of concern

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.


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.


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 $x=y=z=\frac{1}{3}$

  • $\begingroup$ This would only work if you could prove that $\frac{1}{27}\leq xyz$, but you have the opposite inequality. $\endgroup$
    – Arnaud D.
    Commented Jun 20, 2019 at 9:46

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,


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,


which is equal to $-\frac{1}{2}\lbrace a^2(1+2a)+b^2(1+2b)+c^2(1+2c)\rbrace\leq 0$ .

  • $\begingroup$ your last inequality is not true,because $a,b,c\geq -\frac{1}{3}$ so it can happen that some are negative inside bracket .. $\endgroup$
    – Ishan
    Commented May 10, 2020 at 10:36
  • $\begingroup$ $a\geq -\dfrac{1}{3}$ implies $2a\geq -\dfrac{2}{3}$ which implies $1+2a\geq 1-\dfrac{2}{3}=\dfrac{1}{3}>0$ $\endgroup$
    – Debashish
    Commented May 17, 2020 at 3:35

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