It is sufficient to prove $f_{n+1}\leq f_n$ for each $n$.
Let $x \in \Bbb{R}$, then either $x < n$ or $n \leq x < n+1$ or $n+1\leq x.$
If $x
If $n \leq x
if $x \geq n+1$, then $f_{n+1}(x) = \frac{1}{n+1} < \frac{1}{n} = f_n(x).$
It is sufficient to prove $f_{n+1}\leq f_n$ for each $n$.
Let $x \in \Bbb{R}$, then either $x < n$ or $n \leq x < n+1$ or $n+1\leq x.$
If $x
If $n \leq x
if $x \geq n+1$, then $f_{n+1}(x) = \frac{1}{n+1} < \frac{1}{n} = f_n(x).$