Verifying a proof that if $x,y,z \geq 0$ and $x+y+z = 1$, then $0 \le xy + yz + zx - 2xyz \le \frac{7}{27}$

Question

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.

Answer 1

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.

Answer 2

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

Answer 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:

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.

Answer 4

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.

Answer 5

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

Answer 6

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