An inequality by induction: Engel, Problem 24, page 208 I am reading Arthur Engel's Problem Solving Strategies, Section 8, The Induction Principle, Problem 24  with its solution  I do not understand the second inequality in the solution on page 216. This is valid when the sum of the fractions $1/a_i$ is small enough. How do we show that?
Is there an alternative proof?
 A: Note that
$$0<1-\frac{1}{a_1}-\cdots-\frac{1}{a_m}=\frac{a_1\cdots a_m-\widehat{a_1}a_2\cdots a_m-a_1\widehat{a_2}\cdots a_m-\cdots-a_1\cdots\widehat{a_m}}{a_1\cdots a_m}$$
where $\widehat{a_i}$ means that the term of the $i$th index is left out. Since that fraction on the right is positive and the numerator is an integer, the numerator is at least $1$. Therefore
$$1-\frac{1}{a_1}-\cdots-\frac{1}{a_m}\ge \frac{1}{a_1\cdots a_m}$$
which is exactly the second inequality.
The next two steps should be clear (the assumption that $\sum_{i=1}^m 1/a_i=1$ and the induction hypothesis, respectively).
Let us prove the last inequality from that chain in the book:
For $i=m+1,\dots,n$, we define $k$ by setting $i=m+1+k$. Then $0\le k\le n-m-1<n$.
Now estimate
 $i!=(m+1+k)!\ge m!(m+1+k)\ge m!m+m!(1+k)$. Therefore
$$\sum_{i=m+1}^n \frac{1}{2^{i!}}< \sum_{k=0}^{n}\frac{1}{2^{m!m}}\frac{1}{2^{m!(1+k)}}\le\left(\frac{1}{2^{m!}}\right)^m\underbrace{\sum_{k=0}^\infty \frac{1}{2^{m!(1+k)}}}_{\le \sum_{j=1}^\infty \frac{1}{2^j}=1}\le \left(\frac{1}{2^{m!}}\right)^m$$
Taking $m$th roots we get the claimed inequality.
Now altogether we have shown that $a_k\ge 2^{k!}$ for all $k=1,2,\dots,n$. This leads to a contradiction to the assumption $\sum_{i=1}^n \frac{1}{a_i}=1$ because
$$\sum_{i=1}^n \frac{1}{a_i}\le \sum_{i=1}^n \frac{1}{2^{i!}}<\sum_{j=1}^\infty \frac{1}{2^j}=1$$
