(Moscow 2001, problem C3) Find a polynomial of degree 2001 such that $P(x)+P(1-x)=1$ for all $x$.
Let $x=\frac{1}{2}+y$. We are looking for $P$ such that $P\left(\frac{1}{2}+y\right)+P\left(\frac{1}{2}-\right.$ $y)=1$. Let $Q(y)=P\left(\frac{1}{2}+y\right)-\frac{1}{2}$, then $Q(y)+Q(-y)=0$. We can take for example $Q(y)=y^{2001}$, so $P(x)=\frac{1}{2}+\left(x-\frac{1}{2}\right)^{2001}$.

## 2 Thursday afternoon, 21st: Baptiste Louf

## To Start

We begin with a fact that may seem trivial, but on which many inequalities are based: a square is always positive.

Proposition. For all real numbers $a$ and $b$, we have $a^{2}+b^{2} \geq 2 a b$, with equality if and only if $a=b$.

Proof. By subtracting the right-hand side and factoring, we see that the inequality is equivalent to $(a-b)^{2} \geq 0$, which is indeed true, and we note the case of equality.

Corollary (Arithmetic-Geometric Inequality). If $x$ and $y$ are positive real numbers, then $\frac{x+y}{2} \geq \sqrt{x y}$ with equality if and only if $x=y$.

Proof. Apply the above property with $a=\sqrt{x}$ and $b=\sqrt{y}$.