## Task 10/77

Determine all solutions of the equation

$$
\sum_{i=1}^{3} \frac{1}{x_{i}}=1
$$

in natural numbers $x_{i}$, such that $x_{i} \geq x_{j}$ if $i>j$.
Due to $x_{3} \geq x_{2} \geq x_{1}$, we have $x_{1}>1$ and

$$
1=\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}} \leq \frac{1}{x_{1}}+\frac{1}{x_{1}}+\frac{1}{x_{1}}=\frac{3}{x_{1}}
$$

thus $x_{1} \leq 3$. Therefore, two cases are possible:

1. $x_{1}=2$. Then the given equation takes the form

$$
\frac{1}{2}+\frac{1}{x_{2}}+\frac{1}{x_{3}}=1
$$

We further conclude

$$
\frac{1}{2}=\frac{1}{x_{2}}+\frac{1}{x_{3}} \leq \frac{1}{x_{2}}+\frac{1}{x_{2}}=\frac{2}{x_{2}}
$$

from which $x_{2} \leq 4$ follows. This leads to the two solutions $x_{1}=2 ; x_{2}=3 ; x_{3}=6$ and $x_{1}=2 ; x_{2}=4 ; x_{3}=4$. 2. $x_{1}=3$. Analogously, we obtain

$$
\frac{1}{3}+\frac{1}{x_{2}}+\frac{1}{x_{3}}=1
$$

From this, we conclude

$$
\frac{2}{3}=\frac{1}{x_{2}}+\frac{1}{x_{3}} \leq \frac{1}{x_{2}}+\frac{1}{x_{2}}=\frac{2}{x_{2}}
$$

from which $x_{2}=x_{3}=3$ immediately follows. Thus, the third solution is $x_{1}=x_{2}=x_{3}=3$. No further solutions can exist based on the solution method.