TITLE: inequality $\max\{a_1,a_2,\cdots,a_n \}\leq {n^2}^{n-1}.$with Egyptian fraction
QUESTION [5 upvotes]: Let $a_1,a_2,\cdots,a_n $ be positive integer such that$\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}=1.$ Prove that$$\max\{a_1,a_2,\cdots,a_n \}\leq {n^2}^{n-1}.$$
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REPLY [4 votes]: Suppose that $a_1\leq a_2\leq \ldots\leq a_n$. We shall prove that $a_k \leq n^{2^{k-1}}$ for every $k=1,2,\ldots,n$.  First, we have $$\frac{n}{a_1}\geq\sum_{i=1}^n\,\frac{1}{a_i}=1$$
which implies that $$a_1 \leq n=n^{2^{1-1}}\,.$$
Now, assume that the claim holds for $k=1,2,\ldots,r$ for some $r\in\mathbb{N}$ with $r<n$.  Then,
$$\frac{n-r}{a_{r+1}}\geq \sum_{i=r+1}^n\,\frac{1}{a_i}=1-\sum_{i=1}^r\,\frac{1}{a_i}\geq\frac{1}{\prod_{i=1}^r\,a_i}.$$
That is,
$$a_{r+1}\leq (n-r)\,\prod_{i=1}^r\,a_i \leq n\,\prod_{i=1}^r\,n^{2^{i-1}}=n\cdot n^{\sum_{i=1}^r\,2^{i-1}}=n\cdot n^{2^r-1}=n^{2^{(r+1)-1}}\,.$$
By induction, the claim holds.  Thus, $a_n \leq n^{2^{n-1}}$, as required.  In fact, we can improve this inequality and get $a_n\leq \prod_{k=1}^n\,k^{2^{k-2}}\leq n^{2^{n-1}-1}$.