Let $a, b, c$ be nonnegative real numbers. Prove that \[ \frac{a+b+c}{3} - \sqrt[3]{abc} \leq \max\{(\sqrt{a} - \sqrt{b})^2, (\sqrt{b} - \sqrt{c})^2, (\sqrt{c} - \sqrt{a})^2\}. \]
2000 USA Team Selection Test
June 10th - Day 1
Let $ ABCD$ be a cyclic quadrilateral and let $ E$ and $ F$ be the feet of perpendiculars from the intersection of diagonals $ AC$ and $ BD$ to $ AB$ and $ CD$, respectively. Prove that $ EF$ is perpendicular to the line through the midpoints of $ AD$ and $ BC$.
Let $p$ be a prime number. For integers $r, s$ such that $rs(r^2 - s^2)$ is not divisible by $p$, let $f(r, s)$ denote the number of integers $n \in \{1, 2, \ldots, p - 1\}$ such that $\{rn/p\}$ and $\{sn/p\}$ are either both less than $1/2$ or both greater than $1/2$. Prove that there exists $N > 0$ such that for $p \geq N$ and all $r, s$, \[ \left\lceil \frac{p-1}{3} \right\rceil \le f(r, s) \le \left\lfloor \frac{2(p-1)}{3} \right\rfloor. \]
June 11th - Day 2
Let $n$ be a positive integer. Prove that \[ \binom{n}{0}^{-1} + \binom{n}{1}^{-1} + \cdots + \binom{n}{n}^{-1} = \frac{n+1}{2^{n+1}} \left( \frac{2}{1} + \frac{2^2}{2} + \cdots + \frac{2^{n+1}}{n+1} \right). \]
Let $n$ be a positive integer. A $corner$ is a finite set $S$ of ordered $n$-tuples of positive integers such that if $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ are positive integers with $a_k \geq b_k$ for $k = 1, 2, \ldots, n$ and $(a_1, a_2, \ldots, a_n) \in S$, then $(b_1, b_2, \ldots, b_n) \in S$. Prove that among any infinite collection of corners, there exist two corners, one of which is a subset of the other one.
Let $ ABC$ be a triangle inscribed in a circle of radius $ R$, and let $ P$ be a point in the interior of triangle $ ABC$. Prove that \[ \frac {PA}{BC^{2}} + \frac {PB}{CA^{2}} + \frac {PC}{AB^{2}}\ge \frac {1}{R}. \] Alternative formulation: If $ ABC$ is a triangle with sidelengths $ BC=a$, $ CA=b$, $ AB=c$ and circumradius $ R$, and $ P$ is a point inside the triangle $ ABC$, then prove that $ \frac {PA}{a^{2}} + \frac {PB}{b^{2}} + \frac {PC}{c^{2}}\ge \frac {1}{R}$.
These problems are copyright $\copyright$ Mathematical Association of America.