An inequality for two positive series

The question that asks if the proposition must hold for all sufficiently large $m$ must be answered in the negative. I have at the moment no answer for the second question.

Set for $nge 1$

$$a_n=frac1{2^n},$$

and define $i_n in mathbb N$ recursively as

$$i_1=1, i_{n+1}=i_n+frac2{a_{i_n}}, n ge 1$$ and then

$$b_n=frac{a_{i_k}}2, text{ if } i_k le n < i_{k+1}.$$

The $i_n$ sequence is easily seen to tend to $infty$, so the definition of $b_n$ makes sense.

What happens is that $b_1$ starts out as half of $a_1=frac12$, but then $b_n$ stays constant for as long as necessary to reach sum 1, which is 4 for summands for $b_1=frac14$. Then $b_5$ is again half of $a_5=frac1{32}$ and $b_6, b_7,ldots$ are all the same, until again the sum of all $b_n$'s, starting from $b_5$ becomes one (64 summands, so $i_2=5+64=69$).

That means both $(a_n)$ and $(b_n)$ fullfill the conditions of the problem, but still $b_{i_k} = frac{a_{i_k}}2 < a_{i_k}$ for each $k$.

The condition (1) is equivalent to

$$sum_{n=1}^m (b_n-a_n)u_n ge 0$$,

and that means whatever $sum_{n=1}^{i_k-1} (b_n-a_n)u_n$ for any $k$ may be, you can choose $u_{i_k}$ big enough such that $sum_{n=1}^{i_k} (b_n-a_n)u_n < 0$, because the last summand has a negative coefficient $(b_{i_k}-a_{i_k})$.