Lets assume that a prime $p$ is non-lonley then $1+\frac{1}{p}=1+\frac{1}{d_1}+\cdots +\frac{1}{d_n}$which $d_1\cdots d_n$ are divisors of $d_n(\neq p)$ if $p\mid d_n$, there exist $1<i<n$ s.t. $d_i=p$ so it is contradiction with size. else,$\frac{1}{p}=\frac{1}{d_1}+\cdots +\frac{1}{d_n}=\frac{b}{d_1\cdots d_n}$ Hence $pb=d_1\cdots d_n$ so $p\mid LHS, p\nmid RHS$ so contradiction

Let $n=p_1^{e_1}p_2^{e_2}\cdots p_e^{p_e}$ then sum of reciprocals of $n$'s divisors is $(1+\frac{1}{p_1}+\cdots+\frac{1}{p_1^{e_1}})\cdots (1+\frac{1}{p_t}+\cdots +\frac{1}{p_t^{e_t}})$ so it is enough to find two numbers that are non-lonely each other than by multiplying $p^e$($e$is any integer, $p$is prime number that doesn't divide $x,y$) however, every perfect number makes value 2(ex) 6,28) QED