Found problems: 2
a) Given $100$ numbers $a_1, ..., a_{100}$ such that $\begin{cases}
a_1 - 3a_2 + 2a_3 \ge 0, \\
a_2 - 3a_3 + 2a_4 \ge 0, \\
a_3 - 3a_4 + 2a_5 \ge 0, \\
... \\
a_{99} - 3a_{100} + 2a_1 \ge 0, \\
a_{100} - 3a_1 + 2a_2 \ge 0 \end{cases}$
prove that the numbers are equal.
b) Given numbers $a_1=1, ..., a_{100}$ such that $\begin{cases}
a_1 - 4a_2 + 3a_3 \ge 0, \\
a_2 - 4a_3 + 3a_4 \ge 0, \\
a_3 - 4a_4 + 3a_5 \ge 0, \\
... \\
a_{99} - 4a_{100} + 3a_1 \ge 0, \\
a_{100} - 4a_1 + 3a_2 \ge 0 \end{cases}$
Find $a_2, a_3, ... , a_{100}.$
*. Positive numbers $x_1, x_2, ..., x_{100}$ satisfy the system $$\begin{cases} x^2_1+ x^2_2+ ... + x^2_{100} > 10 000 \\
x_1 + x_2 + ...+ x_{100} < 300 \end{cases}$$
Prove that among these numbers there are three whose sum is greater than $100$.