Question:
Let $x_1,$ $x_2,$ $\dots,$ $x_{101}$ be positive real numbers such that $x_1^2 + x_2^2 + \dots + x_{101}^2 = 1.$  Find the maximum value of
\[x_1 x_2 + x_1 x_3 + \dots + x_1 x_{101}.\]

Answer:
By the AM-QM inequality,
\[\frac{x_2 + x_3 + \dots + x_{101}}{100} \le \sqrt{\frac{x_2^2 + x_3^2 + \dots + x_{101}^2}{100}}.\]Then $x_2 + x_3 + \dots + x_{101} \le 10 \sqrt{x_2^2 + x_3^2 + \dots + x_{101}^2},$ so
\[x_1 x_2 + x_1 x_3 + \dots + x_1 x_{101} \le 10x_1 \sqrt{x_2^2 + x_3^2 + \dots + x_{101}^2} = 10x_1 \sqrt{1 - x_1^2}.\]By the AM-GM inequality,
\[x_1 \sqrt{1 - x_1^2} \le \frac{x_1^2 + (1 - x_1^2)}{2} = \frac{1}{2},\]so $10x_1 \sqrt{1 - x_1^2} \le 5.$

Equality occurs when $x_1 = \frac{1}{\sqrt{2}}$ and $x_2 = x_3 = \dots = x_{101} = \frac{1}{10 \sqrt{2}},$ so the maximum value is $\boxed{5}.$