Find the maximum value of $x^3 + y^3 + z^3$ where $x, y, z \in [0, 2]$ and $x + y + z = 3$.

Question

Given that $x, y, z \in [0, 2]$ and $x + y + z = 3$. Calculate the maximum value of $$\large x^3 + y^3 + z^3$$

I'm done. Should you have different solutions, you could post them down below. Having a solution there already, I want to see unfamiliar thoughts.

Answer 1

We have that $x, y, z \le 2 \implies (x - 2)(y - 2)(z - 2) \le 0$

$$\iff xyz - 2(xy + yz + zx) + 4(x + y + z) - 8 \le 0$$

$$\iff 4 \cdot 3 - 2(xy + yz + zx) \le 8 - xyz \implies - 2(xy + yz + zx) \le -(xyz + 4)$$

$$\iff xy + yz + zx \ge \frac{xyz}{2} + 2$$

It is common knowledge that $$x^3 + y^3 + z^3 = (x^3 + y^3 + z^3 - 3xyz) + 3xyz$$

$$ = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) + 3xyz$$

$$ = 3\left[(x + y + z)^2 - 3(xy + yz + zx)\right] + 3xyz \le 3\left[xyz - 3\left(\frac{xyz}{2} + 2\right) + 3^2\right]$$

$$ = 3\left(3 - \frac{xyz}{2}\right) \le 9$$

The equality sig occurs when $\{x, y, z\} = \{0, 1, 2\}$.

Answer 2

By a compactness argument, the maximum is attained. Note that $$(u+h)^3+(v-h)^3=u^3+v^3+3h(u^2-v^2)+3h^2(u+v)=u^3+v^3+3h(u+v)(u-v+h),$$ hence $$ (u+h)^3+(v-h)^3>u^3+v^3$$ if $u\ge v>0$ and $h>0$.

We conclude that for all $(x,y,z)\in[0,2]^2$ with $x+y+z=3$ we can find $(x',y',z')\in[0,2]^3$ with $x'+y'+z'=3$ and $x'^3+y'^3+z'^3>x^3+y^3+z^3$ as long as we find two different values among $x,y,z$ such that both are $\in(0,2)$. Hence at the maximizer, at least two of the values are $\in\{0,2\}$. As $3-2-2<0$ and $3-0-0>2$, the only possibility under the given resrictions is that $x=0,y=1,z=2$ or any permutation thereof.

Answer 3

Let $x\geq y\geq z$.

Thus, $$(2,1,0)\succ(x,y,z).$$ Indeed, $$2\geq x,$$ $$2+1\geq x+y$$ and $$2+1+0=x+y+z.$$ Also $f(x)=x^3$ is a convex functions on $[0,2].$

Thus, by Karamata we obtain: $$2^3+1^3+0^3\geq x^3+y^3+z^3,$$ which says that $9$ is a maximal value.