Question 116
Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$, $f(x) = (x - a_{1})^{2} + (x - a_{2})^{2} + \ldots + (x - a_{2018})^{2}$, where $a_{1}, a_{2}, \ldots, a_{2018}$ are given real numbers.
a) Show that $f(x) \geq f\left(\frac{a_{1} + a_{2} + \ldots + a_{2018}}{2018}\right)$ for every real value of $x$.
b) Show that if $a_{1}^{2} + a_{2}^{2} + \ldots + a_{2018}^{2} = 1$, then $\left|a_{1} + a_{2} + \ldots + a_{2018}\right| \leq \sqrt{2018}$.
a) Show that $f(x) \geq f\left(\frac{a_{1} + a_{2} + \ldots + a_{2018}}{2018}\right)$ for every real value of $x$.
b) Show that if $a_{1}^{2} + a_{2}^{2} + \ldots + a_{2018}^{2} = 1$, then $\left|a_{1} + a_{2} + \ldots + a_{2018}\right| \leq \sqrt{2018}$.