Min problem on integers

Question

Let $n$ be any integer greater than $2^{10^6}$. Given any $s\le (\log_2 n)/1000$ integers $1=q_1\le q_2\le \cdots q_{s-1}\le q_s=n$. Prove that $$\min_\ell\left(\sum_{i=1}^\ell q_i\right)\left(\sum_{i=\ell+1}^s\frac{1}{q_i}\right)\le \frac{1}{64}$$

Answer 1

Let us denote $$\sigma_\ell:=\sum_{i=1}^\ell q_i\qquad\text{and}\qquad\tau_\ell:=\sum_{i=\ell+1}^s\frac{1}{q_i}.$$ Then $$\prod_{\ell=1}^{s-1}\left(\frac{q_\ell}{q_{\ell+1}}\cdot\frac{\sigma_{\ell+1}}{\sigma_\ell}\cdot\frac{\tau_{\ell-1}}{\tau_\ell}\right)=\frac{q_1}{q_s}\cdot\frac{\sigma_{s}}{\sigma_1}\cdot\frac{\tau_0}{\tau_{s-1}}=\sigma_s\tau_0>n,$$ hence there exists $\ell\in\{1,2,\dotsc,s-1\}$ such that $$\frac{q_\ell}{q_{\ell+1}}\cdot\frac{\sigma_{\ell+1}}{\sigma_\ell}\cdot\frac{\tau_{\ell-1}}{\tau_\ell}>n^{1/(s-1)}>n^{1/s}=2^{1000}.$$ On the other hand, the left-hand side equals $$\frac{q_\ell}{q_{\ell+1}}\left(1+\frac{q_{\ell+1}}{\sigma_\ell}\right)\left(1+\frac{1}{q_\ell\tau_\ell}\right) =\frac{q_\ell}{q_{\ell+1}}+\frac{q_\ell}{\sigma_\ell}+\frac{1}{q_{\ell+1}\tau_\ell}+\frac{1}{\sigma_\ell\tau_\ell},$$ so for the same $\ell$ we have $$\frac{q_\ell}{q_{\ell+1}}+\frac{q_\ell}{\sigma_\ell}+\frac{1}{q_{\ell+1}\tau_\ell}+\frac{1}{\sigma_\ell\tau_\ell}>2^{1000}.$$ The first three terms on the left-hand side do not exceed $1$, hence in fact $$\sigma_\ell\tau_\ell<\frac{1}{2^{1000}-3}.$$