Prove that ${1\over x_1}+{1\over x_2}+\dots+{1\over x_n}\lt3$ if no $x_j=10^kx_i+n$ where $x_i,k,n\in\mathbb{Z^+}$ and $x_i\not=x_j$

Question

Prove that ${1\over x_1}+{1\over x_2}+\dots+{1\over x_n}\lt3$ if no $x_j=10^kx_i+n$ where $x_i,k,n\in\mathbb{Z^+}$, $n<10$, and $x_i\not=x_j$

I have attempted this question multiple times and have barely reached anything. I tried to assume WLOG that $x_1\le x_2\le\dots\le x_n$ however I could not continue. I am still new to such inequality questions so any help would be appreciated. Thank you anyways.

Answer 1

As stated this is false:

Taking $n=11$ and $(x_1,\cdots,x_n)=(1,2,3,\cdots, 11)$ then it is clearly true that no $x_j=10^kx_i+11$ (for $k\in \mathbb Z^+$) and $$\sum_{i=1}^{11} \frac 1{i}=3.019877345>3$$