2. the real numbers $a_{1}, a_{2}, \ldots, a_{16}$ fulfill the two conditions

$$
\sum_{i=1}^{16} a_{i}=100 \quad \text { and } \quad \sum_{i=1}^{16} a_{i}^{2}=1000
$$

What is the largest possible value that $a_{16}$ can take?

## Solution
Set $S=a_{1}+\ldots+a_{15}$ and $Q=a_{1}^{2}+\ldots+a_{15}^{2}$. According to AM-QM, $S^{2} \leq 15 Q$. The constraints are $100-a_{16}=S$ and $1000-a_{16}^{2}=Q$. If we square the first and subtract 15 times the second, we get

$$
16 a_{16}^{2}-200 a_{16}-5000=S^{2}-15 Q \leq 0
$$

The quadratic polynomial on the left has the zeros $-25 / 2$ and 25 . Therefore, $a_{16} \leq 25$. If equality applies, then there must also be equality in AM-GM, i.e. $a_{1}=\ldots=a_{15}$. Inserting into the constraints gives the solution $a_{1}=\ldots=a_{15}=5$ and $a_{16}=25$. The largest possible value is therefore indeed $a_{16}=25$.