Sum of harmonic numbers inequality

Question

Suppose $a = a_1+\cdots+a_k$ and $b = b_1+\cdots+b_k$ for some $k\geq 2$, where all variables are positive integers. Can it happen that

$$\frac{1}{a-a_i+1}+\frac{1}{a-a_i+2}+\cdots+\frac{1}{a} \geq \frac{1}{b-b_i}+\frac{1}{b-b_i+1}+\cdots+\frac{1}{b-1}$$ for all $i=1,\ldots,k$?

In the case $a = b$, this can certainly not happen, because we can take $i$ such that $a_i \leq b_i$ (which has to exist because $\sum_i a_i = \sum_i b_i$). Even if $a \neq b$, this seems unlikely to happen. Can we prove it using bounds on the sum of harmonic series?

Answer 1

Estimating both sides* gives $$\log(a/(a-a_i))> \log((b-1)/(b-b_i-1))$$ for $i=1,...,k$.

This is equivalent to $$\frac{b_i+1}{b-1}<\frac{a_i}{a}$$ If we sum both sides from $i=1,...,k$ we get $$1=\sum_{i=1}^k\frac{b_i}{b}<\sum_{i=1}^k\frac{b_i+1}{b-1}<\sum_{i=1}^k\frac{a_i}{a}=1$$ which is a contradiction.

*This is more of a sketch because I have not taken care of the errors with these bounds yet. I will do more in the morning.

I also did a rudimentary computer search which gave some supporting evidence (no counter-example was found).