Maximum Sum of Cubic Products

Our function $f(x,y,z)=x^{3}y^{3}+y^{3}z^{3}+z^{3}x^{3}$ is a symmetric homogeneous degree $6$ polynomial, and we are trying to maximize it on the simplex of side lengths $1$ . The simplex decomposes into one $2$ -dimesional surface, three $1$ -dimensional lines, and three $0$ -dimensional points. We'll look at the interior's of each shape individually. If it has a maximum on the interior of the two dimensional surface, by symmetry it must by where $x=y=z$ . This is because $x=y=z$ is the only point where the gradient vector is perpendicular to the surface, as the normal vector of the surface is the all ones vector. If it has a maximum on any of the one dimensional lines, by a similar symmetry argument, it follows that it must be at center of the line, where the gradient vector of the restricted two dimensional function is perpendicular. This leaves us with three calculations to do. At the point $x=y=z=\frac{1}{3}$ , we get that $f(x,y,z)=3^{-5}$ , at $x=y=\frac{1}{2}$ , $z=0$ , we get $f(x,y,z)=2^{-6}$ , and at the one dimensional point $x=1$ , $z=y=0$ , we have $f(x,y,z)=0$ . Thus the maximum is $\frac{1}{64}$ , so we find that the answer is $65$ .

It is seen that: $x^{3}y^{3}+y^{3}z^{3}+z^{3}x^{3}\leq x^{3}(y+z)^3: "=": z=0$ According to Cauchy Eq: $x^{3}(y+z)^{3}\leq (\frac{x+y+z}{2})^{6}=(\frac{1}{2})^6: "=": x=y+z=\frac{1}{2}$ $=> MAX(x^{3}y^{3}+y^{3}z^{3}+z^{3}x^{3})=(\frac{1}{2})^{6}=\frac{1}{64}=\frac{a}{b}$ $=> a+b=\boxed{65}$

I considered 3 cases:

Case I: None of $x,y$ or $z$ is zero. Note that to maximize the expression, $x=y=z=1/3$ . It will yield $1/243$ .

Case II: Exactly one of $x,y$ or $z$ is equal to zero. By symmetry, we can let $x=0$ . Note that to maximize the expression, $y=z=1/2$ . It will yield $1/64$ .

Case III: Exactly two on $x,y$ or $z$ are zeros. By symmetry, we cal let $x=y=0$ . Note that to maximize the expression, $y=1$ . It will yield 0.

We did not consider the last case $x=y=z=0$ since it will not satisfy the equation.

Thus the maximum value for the expression is $1/64$ . Thus $a=1,b=64$ and $a+b=65$ .

[Latex edits - Calvin]

Derivatives are not applicable and become not relevant as x, y and z are restricted to number between 0 and 1 only; ANY value within boundaries ought to be considered. With relation of x + y + z = 1, search for the wanted expression reduces to 2 dimensions only. By simple computation, maximum and minimum can easily be obtained with adequate fineness, which I really did for no uncertain thought. x, y and z are just to be distributed with values to sum up to 1. But to multiply together and taking a power to 3 will lower down the wanted value. For all symmetries like this, we can turn off some of them but we can only turn off one of them with exact zero. Do not need to take all to use. Therefore taking one of them as zero, the question reduce to x + y = 1 and (x y)^3 wanted. Since a balanced or equal relation between x and y is important for a maximum, x ought to equal to y that x = y = 0.5 and z = 0. I actually obtained simulation result of 0.015625 at a (0, 0.5, 0.5) for example. Therefore, the maximum is (1/ 4)^3 i.e. 1/ 64. Answer: 64 + 1 = 65.

Note: Plus is OR and times is AND in logic.

given,x+Y+z =1,so for getting maximum values ,all the values should be greater than or equal to zero. since z is common in 2 of the terms , for getting maximum value for given expression,z = 0 then the maximum values for (x,y) to satisfy x+y+z = 1, is (0.5,0.5) so,x = 0.5 =1/2, y=0.5=1/2 , z = 0 so,x^3 y^3 +y^3 z^ + z^3 x^3 = (1/2)^3 (1/2)^3 +0+0 =1/64=a/b so,a+b = 1+64=65

If x+y+z=1, then you can use reasoning that a fraction multiplied by a fraction must get smaller. So the largest number would to be putting 1/2 on X and 1/2 on Y and 0 on Z so there will only be one product, X^3*Y^3. That is (1/2)^3(1/2)^3, which is (1/8)(1/8), which is 1/64. 1+64 is 65.

Since x+y+z=1 and x,y,z are more than zero, x,y,z are all less than 1 Hence cube of x,y,z will make them smaller. Since if a<b<c,a^3<b^3<c^3 Hence to obtain the maximum value of (xy)^3 +(yz)^3 +(xz)^3 , One of them must be zero. For example if x is zero,max value of (xy)^3 +(yz)^3 +(xz)^3=(yz)^3 Since y+z=1, to get max value of (yz)^3, y=z=0.5 (yz)^3=1/64, a+b=65

First tried x=y=z=1/3 Then tried x=y=1/2, z=0 so that the first term would be maximal That did it

It is given that $x+y+z = 1$ and $x,y,z > 0$ . For this to be optimized in the domain $0<x,y,z<1$ , one term must be very large and the others can be small. In this case, let $z=0$ . With this, all that is left to optimize is $x^3 y^3$ . For this term to be optimized, $x$ and $y$ have to be very close to each other. So $x=y$ . And, since $x+y+z=1$ , $x=y=\frac{1}{2}$ . Thus, the optimized value of this polynomial is $\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^3+0+0=\frac{1}{64}$ . $1+64=65$ .