(IMO 2016, problem 5)

We write on the board the equality:

$$
(x-1)(x-2) \ldots(x-2016)=(x-1)(x-2) \ldots(x-2016)
$$

We want to erase some of the 4032 factors in such a way that the equation on the board has no real solutions. What is the minimum number of factors that must be erased to achieve this?

## - Solutions -
If there remains a factor $x-1$ on the right and one on the left, then the equation admits 1 as a solution, and the same applies to the other factors. Therefore, at least one $x-1$ must be erased, at least one $x-2$, etc., which means at least 2016 factors must be erased.

It remains to show that it is possible to do so by erasing exactly 2016 numbers. One way to do this is to erase $x-i$ on the left if $i$ is congruent to 0 or 1 modulo 4, and on the right if $i$ is congruent to 2 or 3 modulo 4.

## 4 Inequalities

The course covered the following concepts:

- Arithmetic-geometric inequality
- Turnstile lemma: $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$
- Cauchy-Schwarz inequality (proof using the discriminant of a well-chosen polynomial, equality case)
- Inequality of bad students (proof using Cauchy-Schwarz)
- Convexity: definitions, properties, Jensen's inequality, and proof of the AGI

These concepts can be found in the very comprehensive course notes by Pierre Bornsztein on the subject.

## - Exercises -