Task A-1.4. (4 points)

Let $a, b$ and $c$ be real numbers such that

$$
a+b+c=3 \quad \text { and } \quad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0
$$

What is $a^{2}+b^{2}+c^{2}$?
## Solution.

Since $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$

from the first given equality, we get

$$
\begin{aligned}
& a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a=9 \\
& a^{2}+b^{2}+c^{2}=9-2(a b+b c+c a)
\end{aligned}
$$

Since from the second given equality it follows that

$$
\frac{b c+c a+a b}{a b c}=0
$$

which means $a b+b c+c a=0$,

we finally get:

$$
\begin{aligned}
a^{2}+b^{2}+c^{2} & =9-2(a b+b c+c a) \\
& =9-2 \cdot 0=9
\end{aligned}
$$