min(x², (a-x)²) ≤ a²/4

Dependencies: None

Let $a \in \mathbb{R}_{\ge 0}$ and $f(x) = \min(x^2, (a-x)^2)$. Then $f(x) ≤ a^2/4$ for all $x \in [0, a]$ and $f(a/2) = a^2/4$.

Proof

If $x ≤ a/2$, then $x^2 ≤ a^2/4$. If $x ≥ a/2$, then $(a-x)^2 ≤ a^2/4$. Hence, $f(x) ≤ a^2/4$.

Dependency for:

1. PROP cake division doesn't exist for supermodular valuations

Info:

• Depth: 0
• Number of transitive dependencies: 0

Transitive dependencies: None