## Task A-2.2.

Determine all triples $(a, b, c)$ of real numbers for which

$$
a^{2}+b^{2}+c^{2}=1 \quad \text { and } \quad(2 b-2 a-c) a \geqslant \frac{1}{2}
$$
## First Solution.

Let $a, b$, and $c$ be such that the given conditions are satisfied.

Then we have

$$
(2 b-2 a-c) a \geqslant \frac{a^{2}+b^{2}+c^{2}}{2}
$$

By multiplying the inequality by 2 and moving terms to one side, we get

$$
0 \geqslant a^{2}+b^{2}+c^{2}-4 a b+4 a^{2}+2 a c
$$

It follows that

$$
0 \geqslant a^{2}+2 a c+c^{2}+b^{2}-4 a b+4 a^{2}
$$

or

$$
0 \geqslant(a+c)^{2}+(b-2 a)^{2}
$$

Since $(a+c)^{2} \geqslant 0$ and $(b-2 a)^{2} \geqslant 0$, we have

$$
0 \leqslant(a+c)^{2}+(b-2 a)^{2}
$$

Thus, $(a+c)^{2}+(b-2 a)^{2}=0$, which implies $c=-a$ and $b=2 a$

From the condition $a^{2}+b^{2}+c^{2}=1$, we get

$$
a^{2}+4 a^{2}+a^{2}=1, \quad \text { i.e. } \quad a^{2}=\frac{1}{6}
$$

The desired triples are: $(a, b, c)=\left(\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}}\right)$ and $(a, b, c)=\left(-\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)$.