2. Problem: Find all triples $(a, b, c)$ of real numbers such that
$$
a^{2}+b^{2}+c^{2}=1 \quad \text { and } \quad a(2 b-2 a-c) \geq \frac{1}{2}
$$
Solution: (Viet Hoang)
The equations can be rewritten as
$$
a^{2}+b^{2}+c^{2}=1 \quad \text { and } \quad 4 a b-4 a^{2}-2 a c \geq 1 .
$$

Substituting the first equation into the second and rearranging yields
$$
\begin{aligned}
4 a b-4 a^{2}-2 a c & \geq a^{2}+b^{2}+c^{2} \\
5 a^{2}+b^{2}+c^{2}+2 a c-4 a b & \leq 0 \\
(2 a-b)^{2}+(a+c)^{2} & \leq 0 .
\end{aligned}
$$

Hence $b=2 a$ and $c=-a$. But $a^{2}+b^{2}+c^{2}=1$. So
$$
a^{2}+(2 a)^{2}+(-a)^{2}=1
$$
and thus $a= \pm \frac{1}{\sqrt{6}}$. Therefore the final answers are:
$$
(a, b, c)=\left(\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}\right) \text { and }\left(\frac{-1}{\sqrt{6}}, \frac{-2}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right) \text {. }
$$