Problem

Let $x,y,z\in [0,1]$, prove that

$$ 2(x^3+y^3+z^3)-x^2y-y^2z-z^2x\leq 3. $$

Solution - by Ammar Ratnani (Stanford)

As stated in the problem, assume that $$x, y, z \in [0,1]$$. Also define

$$ L(x, y, z) = 2 \cdot (x^3 + y^3 + z^3) - x^2 y - y^2 z - z^2 x. $$

to be the left-hand side of the inequality in the problem. The domain of $L$ is $[0,1]^3$. The goal is to show that $L(1, 1, 1) = 3$ is in fact the maximum value $L$ takes on that domain.

Observation

The value of $L$ is stays the same under cyclic shifts of its arguments. In other words, $L(x, y, z) = L(y, z, x) = L(z, x, y)$.

Proof

Using commutativity of addition, observe that

$$ \begin{align*} L(x, y, z) &= 2 \cdot (x^3 + y^3 + z^3) - x^2 y - y^2 z - z^2 x) \\ &= 2 \cdot (y^3 + z^3 + x^3) - y^2 z - z^2 x - x^2 y) \\ &= L(y, z, x). \end{align*} $$

Apply the same argument on $L(y, z, x)$ to complete the proof of the observation.

Because of this symmetry, we can focus on maximizing $L$ over one coordinate --- say $x$. Then, we can extend our result to all three coordinates using these cyclic shifts.

Proposition

For all $x, y, z \in [0,1]$, $L(x, y, z) \leq L(x, y, 1)$.

Proof

Expanding both sides and removing like terms reveals that proving this is equivalent to showing that

$$ \begin{align*} 2 z^3 - y^2 z - z^2 x &\leq 2 - y^2 - x \\ y^2 \cdot (1 - z) + x \cdot (1 - z^2) &\leq 2 \cdot (1 - z^3). \end{align*} $$

Now since $z \in [0, 1]$, $z \geq z^2 \geq z^3$, which implies $1 - z \leq 1 - z^2 \leq 1 - z^3.$ Furthermore, these inequalities remain valid when multiplied by any non-negative number. As a result,

$$ \begin{align*} y^2 \cdot (1 - z) &\leq y^2 \cdot (1 - z^3) \\ x \cdot (1 - z^2) &\leq x \cdot (1 - z^3), \end{align*} $$

and summing those gives

$$ y^2 \cdot (1 - z) + x \cdot (1 - z^2) \leq y^2 \cdot (1 - z^3) + x \cdot (1 - z^3). $$