On a $(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)=2 a b+2 a c+2 b c=4^{2}-12=4$. Hence $a b+a c+b c=2$. Thus $a b c \times\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2$ and therefore $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2$ since $a b c=1$.

The arithmetic mean of $a, b$, and $c$ is thus $\frac{a+b+c}{3}=\frac{4}{3}$.

Similarly, their harmonic mean is $\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{3}{2}$.

Thus, the inequality between the arithmetic mean and the harmonic mean is not satisfied, so at least one of the numbers $a, b$, or $c$ is not a positive real number (and therefore a second one is not either, since their product is $1>0$).

## 3 Group C: Polynomials

## 1 Thursday 20 morning: Igor Kortchemski

This was a course on polynomials in one variable. The following concepts were covered: operations on polynomials, Euclidean division of polynomials, roots, factorization, multiple roots and the derivative polynomial, interpolation, (Lagrange polynomials), elementary symmetric polynomials, Viète's formulas, Newton's formulas.

## Exercises given in class