Inequality for Fibonacci sequence

Let $\{f_n\}_{(n\ge 1)}$ Fibonacci sequence, which satisfies the following recurrence relation;
$$f_1=f_2=1, f_{n+2}=f_n+f_{n+1} \phantom{a} (n\ge 1)$$
A few months ago I found an interesting inequality;
$$\forall N\in\mathbb{N};\phantom{a}\sum_{n=1}^{N}\frac{f_n}{f_{n+1}}>\frac{-1+\sqrt{5}}{2}N\quad\cdots (1)$$
This can be proved amazingly simple;

Let $\displaystyle S_N=\sum_{n=1}^{N}\frac{f_n}{f_{n+1}}$ and $\displaystyle T_N=\sum_{n=1}^{N}\frac{f_{n+1}}{f_n}$. By Cauchy-Schwarz inequality $S_N T_N> N^2$. Using the given recurrence relation we can easily prove that $T_NT_N S_N>N^2$, so $S_N^2+NS_N-N^2>0$. Solving this quadratic inequality, we get $\displaystyle S_N>\frac{-1+\sqrt{5}}{2}N$.

After proving (1) I tried to find the upper bound of the sum by using similar method. I finally found
$$\forall N\in\mathbb{N}; \phantom{a}\sum_{n=1}^{N}\frac{f_{n+1}}{f_n}<\frac{1+\sqrt{5}}{2}N+\sum_{n=1}^{N}\frac{1}{n}\quad\cdots (2)$$
However, my proof of (2) is long and complex than I expected; Is there any method to obtain (better) upper bound by using simple method like the proof of (1)? Let's share some ideas.

Answer

HINT A stronger statement is in fact true. $$\dfrac{f_{n+1}}{f_n} < \dfrac{1+\sqrt{5}}2 + \dfrac1n$$
and
$$\dfrac{f_{n}}{f_{n+1}} < \dfrac{\sqrt{5}-1}2 + \dfrac1{n+1}$$