IMO 1996 SL A5

Origin: ROM | Category: Algebra

Problem

Let $P(x)$ be the real polynomial function

$P(x) = ax^3 + bx^2 + cx + d.$

Prove that if

$|P(x)| \leq 1 \quad \text{for all } x \text{ such that } |x| \leq 1,$

then

$|a| + |b| + |c| + |d| \leq 7.$

Solution

Considering the polynomials $\pm P(\pm x)$ we may assume w.l.o.g. that $a, b \geq 0$. We have four cases:

1. $c \geq 0$, $d \geq 0$. Then

$|a| + |b| + |c| + |d| = a + b + c + d = P(1) \leq 1.$ 2. $c \geq 0$, $d < 0$. Then

$|a| + |b| + |c| + |d| = a + b + c - d = P(1) - 2P(0) \leq 3.$ 3. $c < 0$, $d \geq 0$. Then

\begin{align*}

|a| + |b| + |c| + |d| &= a + b - c + d \

&= \frac{4}{3}P(1) - \frac{1}{3}P(-1) - \frac{8}{3}P\left(\frac{1}{2}\right) + \frac{8}{3}P\left(-\frac{1}{2}\right) \leq 7.

\end{align*} 4. $c < 0$, $d < 0$. Then

\begin{align*}

|a| + |b| + |c| + |d| &= a + b - c - d \

&= \frac{5}{3}P(1) - 4P\left(\frac{1}{2}\right) + \frac{4}{3}P\left(-\frac{1}{2}\right) \leq 7.

\end{align*}

Remark. It can be shown that the maximum of $7$ is attained only for

$P(x) = \pm(4x^3 - 3x).$