59 Dados ocho números reales $a_1 \leq a_2 \leq \cdots \leq a_7 \leq a_8$ . Sea $x = \frac{ a_1 + a_2 + \cdots + a_7 + a_8}{8}$ , $y = \frac{ a_1^2 + a_2^2 + \cdots + a_7^2 + a_8^2}{8}$ . Demuestra que \[2 \sqrt{y-x^2} \leq a_8 - a_1 \leq 4 \sqrt{y-x^2}.\] Amir