
Problem 1. Find all triples $(a, b, c)$ of real numbers such that the following system holds:

$$
\left\{\begin{array}{l}
a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\
a^{2}+b^{2}+c^{2}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}
\end{array}\right.
$$


Solution. First of all if $(a, b, c)$ is a solution of the system then also $(-a,-b,-c)$ is a solution. Hence we can suppose that $a b c>0$. From the first condition we have

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

Now, from the first condition and the second condition we get

$$
(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2}-\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)
$$

The last one simplifies to

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

First we show that $a+b+c$ and $a b+b c+c a$ are different from 0 . Suppose on contrary then from relation (1) or (2) we have $a+b+c=a b+b c+c a=0$. But then we would have

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

which means that $a=b=c=0$. This is not possible since $a, b, c$ should be different from 0 . Now multiplying (1) and (2) we have

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

Since $a+b+c$ and $a b+b c+c a$ are different from 0 , we get $(a b c)^{2}=1$ and using the fact that $a b c>0$ we obtain that $a b c=1$. So relations (1) and (2) transform to

$$
a+b+c=a b+b c+c a .
$$

Therefore,

$$
(a-1)(b-1)(c-1)=a b c-a b-b c-c a+a+b+c-1=0 \text {. }
$$

This means that at least one of the numbers $a, b, c$ is equal to 1 . Suppose that $c=1$ then relations (1) and (2) transform to $a+b+1=a b+a+b \Rightarrow a b=1$. Taking $a=t$ then we have $b=\frac{1}{t}$. We can now verify that any triple $(a, b, c)=\left(t, \frac{1}{t}, 1\right)$ satisfies both conditions. $t \in \mathbb{R} \backslash\{0\}$. From the initial observation any triple $(a, b, c)=\left(t, \frac{1}{t},-1\right)$ satisfies both conditions. $t \in \mathbb{R} \backslash\{0\}$. So, all triples that satisfy both conditions are $(a, b, c)=\left(t, \frac{1}{t}, 1\right),\left(t, \frac{1}{t},-1\right)$ and all permutations for any $t \in \mathbb{R} \backslash\{0\}$.

Comment by PSC. After finding that $a b c=1$ and

$$
a+b+c=a b+b c+c a
$$

we can avoid the trick considering $(a-1)(b-1)(c-1)$ as follows. By the Vieta's relations we have that $a, b, c$ are roots of the polynomial

$$
P(x)=x^{3}-s x^{2}+s x-1
$$

which has one root equal to 1 . Then, we can conclude as in the above solution.
