How Many Solutions are Available?

At first, I am going to show that at least one of the variables is less than $4$ . If all the variables equal to $4$ , then we would have:

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \le \frac{1}{4} +\frac{1}{4} +\frac{1}{4} =\frac{3}{4}$ .

Assuming $x \le y \le z$ ,

$x$ has two possible values $x = 2 \text { or } x =3.$

case 1: If $x=2 \implies \frac{1}{y}+\frac{1}{z} = 1-\frac{1}{x} = \frac{1}{2}\implies \frac{1}{y}+\frac{1}{z}-\frac{1}{2}=0\implies$

$yz - 2y -2z = 0\\ yz -2y -2z +4 = 4\\ \text { or } (y-2)(z-2) = 4$

since $y$ and $z$ can't be less than $1$ , neither $(y-2) \text { nor } ( z-2)$ can be negative, the only possible cases are available:

$y-2= 2, z-2 = 2 \implies y=4, z=4 .\\ y-2=1 , z-2=4 \implies y =3, z=6.$

case 2: If $x=3$ , there is only one possibility here:

$2y - 3 = 3, 2z-3 = 3 \implies y=3, z= 3$ .

Thus, the only solutions for the sum of the reciprocals of three natural number to be $1$ is in this form:

$\frac{1}{2}+\frac{1}{4}+\frac{1}{4} =1\\\frac{1}{2}+\frac{1}{3}+\frac{1}{6} =1\\\frac{1}{3}+\frac{1}{3}+\frac{1}{3} =1$