# Upper bound for sum with $\frac{b_i-a_i}{b_i-a_1}$

Let $0<a_1,\dots,a_n<1\leq b_1,\dots,b_n$. For $1\leq i\leq n$, define $$M_i=\frac{1}{\dfrac{b_i-a_i}{b_i-a_1}+\dfrac{b_i-a_i}{b_i-a_2}+\dots+\dfrac{b_i-a_i}{b_i-a_n}}.$$ Let $M=M_1+\dots+M_n$. Is $M$ bounded above by some constant (independent of $n,a_1,\dots,a_n,b_1,\dots,b_n$)?

If we assume (without loss of generality) that $a_1\geq\dots\geq a_n$, we get $M_i\leq\frac{1}{i}$, so $M\leq 1+\frac12+\dots+\frac1n$. But this sum is not bounded by a constant.

• $n$ is fixed in each case and the upper bound should be intended about quantities $a_i$ and $b_i$ for $i=1,2,...,n$ – Piquito Sep 1 '16 at 14:42

A tough nut to crack. But the upper bound you give is in fact optimal. So there is no finite upper bound, uniform in $n$:
Fix $n$, let $\Lambda>1$ and pick numbers $$a_i=1-\Lambda^{-2i} \ \ \mbox{and} \ \ b_i=1+\Lambda^{-2i+1}$$ Then $$\frac{b_i-a_i}{b_i-a_j} = \frac{\Lambda +1}{\Lambda + \Lambda^{2(i-j)}}$$ When $j<i$ we have $$0 \leq \frac{b_i-a_i}{b_i-a_j} \leq \frac{\Lambda+1}{\Lambda+\Lambda^2} \leq \frac{1}{\Lambda}$$ while for $j\geq i$ $$1\leq \frac{b_i-a_i}{b_i-a_j} \leq 1+\frac{1}{\Lambda}$$ Therefore $$n-i+1 \leq \frac{1}{M_i}\leq n-i+1 + \frac{n}{\Lambda}$$ Letting $\Lambda$ go to infinity we obtain: $$\lim_{\Lambda\rightarrow \infty} \ \sum_i {M_i} = \sum_{k=1}^n \frac{1}{k}$$ thus showing that the stated bound is optimal.