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

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.

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.