A Simple Question About Min and Max Operators on Fractional Expression

Question

Let $f: \mathbb{N} \to (0,\infty)$ be some function. I am wondering if the following equality holds $$ \min_{0 \leq k \leq N} \frac{f(k)}{\max_{i\leq k}f(i)} =?= \min_{0\leq \ell \leq k \leq N} \frac{f(k)}{f(\ell)} $$ I believe it is, but fail to see how to prove it...I think $``\geq"$ is trivial but the other direction is not very clear to me. I guess $$ \min_{0 \leq k \leq N} \frac{f(k)}{\max_{i\leq k}f(i)} \leq \min_{0 \leq k \leq N} \frac{f(k)}{f(i)} , \forall i \leq k $$ Then I get stuck to pursue further. Any suggestion is appreciated.

Answer 1

$\;\;\;\;$ Let $\frac{f(k_0)}{f(\ell_0)}=\text{min}_{0\leq k\leq N}\frac{f(k)}{\text{max}_{i\leq k} f(i)}$ where $\ell_0\leq k_0$ so that $\text{min}_{0\leq\ell\leq k\leq N}\frac{f(k)}{f(\ell)}\leq\frac{f(k_0)}{f(\ell_0)}$. On the other hand, letting $\frac{f(k_1)}{f(\ell_1)}=\text{min}_{0\leq\ell\leq k\leq N}\frac{f(k)}{f(\ell)}$ with $\ell_1\leq k_1$, we must have $\frac{f(k_0)}{f(\ell_0)}\leq\frac{f(k_1)}{\text{max}_{i\leq k_1}\;f(i)}\leq\frac{f(k_1)}{f(\ell_1)}$ which finally implies $$\frac{f(k_0)}{f(\ell_0)}=\frac{f(k_1)}{f(\ell_1)}$$ i.e., $\text{min}_{0\leq k\leq N}\frac{f(k)}{\text{max}_{i\leq k} f(i)}=\text{min}_{0\leq\ell\leq k\leq N}\frac{f(k)}{f(\ell)}$.