Let $a,b,c\in\mathbb{N}$ and $a<b<c$. Please find all the triples for which: $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1,$$ using the pigeon hole principle.
Lemma 1:
$a=2$
Proof: Suppose $a=1$, because $\frac{1}{n}>0, \forall n \in \mathbb{N}$, its clear that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>1.$, therefore $a\neq1$.
Suppose $2<a$, if we look at the reciprocal then $\frac{1}{2}>\frac{1}{a}$ and we can add $b,c$ into the mix like so: \begin{align} \frac{1}{2}+&\frac{1}{b}+\frac{1}{c}>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\\ \Rightarrow \frac{1}{2}+&\frac{1}{b}+\frac{1}{c}>1\\ \Rightarrow &\frac{1}{b}+\frac{1}{c}>\frac{1}{2}. \end{align} From the inequalities above, $2<a$ we can say $2<3\leq a$, which tells us $a< 4\leq b$, and $b < 5 \leq c$. We can conclude from this that $4\leq b$ and $5\leq c$ which says: \begin{align} \frac{1}{4}+\frac{1}{5}\geq \frac{1}{b}+\frac{1}{c} &>\frac{1}{2}\\ \Rightarrow \frac{9}{20}&>\frac{1}{2}. \end{align} Which is a contradiction. qed
Conclusion
So I know the only solution to this is $(a,b,c)=(2,3,6)$, but this came from a book on the pigeon hole principle and I just cant see how. I also know I can make a brute force solution show that anything else leads to a contradiction.
With $a=2$ we can change the problem to: $$ \frac{1}{b}+\frac{1}{c} = \frac{1}{2}. $$ Where $b<c$, show that there is one solution using the pigeon hole principle.
Thanks