Prove maximum value of $(z-xy)(x-yz)(y-zx)$ is $\frac{1}{64}$ given $x,y,z \in (0,1)$

I can make it $\frac{1}{64}$ by setting $x,y,z = \frac{1}{2}$, but I have no idea how to show that's the maximum.

  • $\begingroup$ Be careful with your notation for intervals. The notation $(-1,\frac{1}{64})$ means the open interval from $-1$ to $\frac{1}{64}$, i.e. the set $\{x \in \mathbb{R} : -1 < x < \frac{1}{64}\}$. So this set does not contain the elements $-1$ and $\frac{1}{64}$. If you want to denote the closed interval $\{x \in \mathbb{R} : -1 \leq x \leq \frac{1}{64}\}$, use square brackets, i.e. $[-1,\frac{1}{64}]$. $\endgroup$
    – Arthur
    Oct 12, 2013 at 4:16
  • $\begingroup$ I shall make the appropriate change to the question, since I only really want the maximum. $\endgroup$ Oct 12, 2013 at 4:18
  • $\begingroup$ The set of maxima is invariant under permutations of the variables, so if you can show that the max is unique, then it must occur at $(1/2,1/2,1/2)$. $\endgroup$ Oct 12, 2013 at 4:41
  • $\begingroup$ @PrahladVaidyanathan I must apologise for not knowing what that means. I think you're saying that each terms must be the same because they are of the same form. I also have no idea how to show the max is unique. $\endgroup$ Oct 12, 2013 at 4:45
  • $\begingroup$ @PrahladVaidyanathan No, I've got it. You're saying that if there are more than one maximum, there will $3n$ or $3n+1$ of them. $\endgroup$ Oct 12, 2013 at 4:51

4 Answers 4


As it happens, just yesterday I showed on this forum, in the process of answering another stackexchange question, that if we fix the average $a=\frac13(x+y+z)$ then $$ \left( 1 - \frac{xy}{z} \right) \left( 1 - \frac{yz}{x} \right) \left( 1 - \frac{xz}{y} \right) \leq (1-a)^3 $$ with equality iff $x=y=z=a$. Also, by the inequality on arithmetic and geometric means (which also figured in the proof of the $(1-a)^3$ bound), we have $xyz \leq a^3$, again with equality iff $x=y=z=a$. Multiplying these two inequalities yields $$ (z - xy) (x - yz) (y - xz) \leq a^3 (1-a)^3 = \bigl( a(1-a) \bigr)^3, $$ and one final application of the AM-GM inequality shows that this is at most $(1/4)^3 = 1/64$ with equality iff a=1/2, QED.

  • $\begingroup$ This is in fact the end of my simpler proof to that question, but I really don't understand that proof at all. I don't understand why we initially fix the average, for instance. $\endgroup$ Oct 12, 2013 at 14:21
  • $\begingroup$ It's a reasonably common technique. The idea is you're trying to show the maximum occurs when all the variables are equal, and often this can be done by showing the function gets larger when all variables are replaced by their average. $\endgroup$ Oct 12, 2013 at 14:26
  • $\begingroup$ +1 Nice. I tried showing that it was $\leq xyz(1-x)(1-y)(1-z)$ but got stuck. $\endgroup$
    – Calvin Lin
    Oct 12, 2013 at 14:35
  • $\begingroup$ I think I'd need it fleshing out a bit more to understand the hows and whys. $\endgroup$ Oct 12, 2013 at 16:36
  • $\begingroup$ @stevemarvell: this is indeed a special case of general "mixing variables" technique. You can look at these notes to see how it works in practice all just google to find better source. $\endgroup$
    – leshik
    Oct 12, 2013 at 17:43

Here is a straightforward approach. Let us put $f(x,y,z) := (-xy+z)(-yz+x)(-xz+y). $ Now we find its critical points as the solutions of the polynomial system $$\{-2y^2z^2x+y^3z+yz^3+3yzx^2-2y^2x-2z^2x+yz=0,$$ $$-2yz^2x^2+3y^2zx+z^3x+zx^3-2yz^2-2yx^2+zx=0,$$ $$-2y^2zx^2+y^3x+3yz^2x+yx^3-2y^2z-2zx^2+yx=0 \}\,(1)$$ which consists of the partial derivatives of $f$ equating to 0. Making use of the resultant, we eliminate $x$ from the system with help of Maple (Of course, it can be done by hand too.): $$ resultant(-2*y^2*z^2*x+y^3*z+y*z^3+3*y*z*x^2-2*y^2*x-2*z^2*x+y*z, $$ $$ 2*x^2*y*z^2+x^3*z+3*x*y^2*z+x*z^3-2*x^2*y-2*y*z^2+x*z, x)$$ outputs $$ 4y z^3 ( 16\,y^8z^4+8\,y^6z^6+8\,y^4z^8+16\,y^8z^2+$$ $$36\,y^6z^4+25y^4z^6+7y^2 z^8-4y^8-$$ $$24\,y^6{z}^{2}+22\,{y}^{4}{z}^{4}-7\,{y}^{2}{z }^{6}+{z}^{8}+4\,{y}^{6}-$$ $$7\,{y}^{4}{z}^{2}+9\,{y}^{2}{z}^{4}-$$ $$\left.2\,{z}^{6 }-{y}^{2}{z}^{2}+{z}^{4} \right) \,\,(2) $$ and $$resultant(-2*y*z^2*x^2+3*y^2*z*x+z^3*x+z*x^3-2*y*z^2-2*y*x^2+z*x,$$ $$2*x^2*y^2*z+x^3*y+x*y^3+3*x*y*z^2-2*x^2*z-2*y^2*z+x*y , x) $$ outputs $$ 16\,{y}^{2}{z}^{2} \left( {y}^{2}-{z}^{2} \right) ^{3} \left( {y}^{2}{ z}^{2}-{y}^{2}-{z}^{2}+1 \right)=$$ $$16y^2z^2 ( y^2-z^2)^3(z-1)(z+1)(y-1)(y+1)\,\,(3) .$$ Now we can find all the real solutions of equation (3) satisfying the constraints $y\ge 0, y \le 1, z\ge 0,z \le 1$: $$ [\{y = y, z = 0\}, \{y = y, z = 1\},\{y = 0, z = z\}, \{y = 1, z = z\}, \{y = z, z = z\}].$$ All the real solutions of system (1) satisfying the constraints are $$\{x = 0, y = 0, z = z\}, \{x = 0, y = y, z = 0\}, \{x = x, y = 0, z = 0\}, \{x = x, y = x, z = 1\}, \{x = 1, y = y, z = y\},\{x = x, y = 1, z = x\},\{x = 1/2, y = 1/2, z = 1/2\} .$$ The Hessian of $f$ is negative definite only at the point $\{x = 1/2, y = 1/2, z = 1/2\} $ which is the only maximum point of $f$ in the cube $\{0<x,x<1,0<y,y<1,0<z,z<1\}.$

It remains to consider $f$ on the faces of the unit cube. For example, putting $x=0$, we obtain the restriction of $f$ which equals $ -y^2z^2 $ so it is nonpositive on the square $\{y \ge 0,y \le 1,z \ge 0, z\le 1\}.$ Putting $x=1$, we have the restriction of $f$ equals $\left( -y+z \right) \left( -yz+1 \right) \left( y-z \right)$. This also is nonpositive ibid. The behavior of $f$ on the four others is the same. Therefore, we draw the conclusion the maximum of $f$ on the unit cube $\{0<x,x<1,0<y,y<1,0<z,z<1\}$ (which equals $\frac 1 {64}$) is attained at $\{x = 1/2, y = 1/2, z = 1/2\} $ . See the calculations done with Maple here as a PDF file.

  • $\begingroup$ I find it remarkable that the Mathematica command $$Maximize[\{(z - x*y)*(x - y*z)*(y - z*x), 0 <= x, x <= 1, y >= 0, y <= 1, z >= 0, z <= 1\}, \{x, y, z\}] $$ produces $$ \{1/64, \{x -> 1/2, y -> 1/2, z -> 1/2\}\}, $$ mechanizing the answer done by hand. $\endgroup$
    – user64494
    Oct 12, 2013 at 15:54
  • $\begingroup$ Did it show its working out? :) $\endgroup$ Oct 12, 2013 at 21:38

Here is a cheap way. Denote the LHS by $P$. First note that all three factors must be positive to get a positive product.

Now write $(x-yz)(y-xz)=xy(1+z^2)-(x^2+y^2)z\le xy(1+z^2)-2xyz=xy(1-z)^2$. Multiply by $z-xy$ to get $$ P\le xy(1-z)^2(z-xy)\,. $$ Since everything is symmetric, we can just as well write $$ P\le yz(1-x)^2(x-yz) $$ and $$ P\le xz(1-y)^2(y-xz)\,. $$ Multiplying these out, we get $$ P^3\le [xyz(1-x)(1-y)(1-z)]^2P\, $$ so $P\le xyz(1-x)(1-y)(1-z)$. However, $x(1-x)=\frac 14-(x-\frac 12)^2\le\frac 14$ and the same is true for the other two products.

  • $\begingroup$ Is it not possible that two of them could be negative? $\endgroup$ Oct 12, 2013 at 20:40
  • 1
    $\begingroup$ @stevemarvell If $x-yz,y-xz<0$, then $x<yz\le y$ and $y<xz\le x$ simultaneously. $\endgroup$
    – fedja
    Oct 12, 2013 at 20:49
  • $\begingroup$ I guess not then :) $\endgroup$ Oct 12, 2013 at 21:32

Well you can think it in this way. Let's suppose it has a maximum (the problem ask to find it, so I think we can safely assume it has one). Now give the symmetry of the function is easy to see that if you change any of the 3 variables the function does not change. Geometrically it means you could change and of the 3 axis and the graph would not change. That means that the maximum must be at $x=y=z$. Otherwise you could exchange two axis and the graph would be changed (rotated) and the maximum would be shifted (think in 2 dimension to an oval and try to see what happens if you exchange $x$ with $y$), but the function will not change so this is not possible. Now given the fact that the $x,y,z$ are equal let's call them $x=y=z=s$, our equation will become $$ (s-s^2)^3=\frac{1}{64} $$ that means $$ s-s^2=1/4 $$ and therefore given the conditions $s=1/2$. I hope is understandable.

  • $\begingroup$ @ Umberto: You wrote:"That means that the maximum must be at $x=y=z$" . Could you base this claim in detail? $\endgroup$
    – user64494
    Oct 12, 2013 at 19:58
  • $\begingroup$ I think an original comment suggests that if you can prove that there is only one maxima then due to the invariance due to permutation of the terms, they must all be the same. $\endgroup$ Oct 12, 2013 at 20:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .