Since $\frac1z=\int\limits_0^1s^{z-1}\mathrm ds$ for every $z\gt0$, $ f_n(x)=\frac1n\sum_{k=0}^{n-1}\int_0^1s^{k/n}\cdot s^{x-1}\mathrm ds=\int_0^1\frac{1-s}{h_n(s)}\cdot s^{x-1}\mathrm ds, $ with $ h_n(s)=n(1-s^{1/n}). $ For every $c$, the derivative of the function $u\mapsto(1-\mathrm e^{-cu})/u$ is proportional to $1+cu-\mathrm e^{cu}\leqslant0$. Applying this to $u=1/n$ and $c=-\log(s)\geqslant0$ for every $s$ in $(0,1)$, one sees that the sequence $(h_n)_n$ is nondecreasing, hence the sequence $(f_n)_n$ is nonincreasing.
Now, for every $n$, $f_{n+1}(x_n)\leqslant f_n(x_n)=1=f_{n+1}(x_{n+1})$ and the function $f_{n+1}$ is nonincreasing hence $x_{n+1}\leqslant x_n$.
Likewise, consider the functions $ g_n(y)=\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{y+\frac1{2n}+\frac{k}{n}}. $ The integral representation used above nowyields $ g_n(y)=\int_0^1\frac{1-s}{k_n(s)}\cdot s^{y-1}\mathrm ds, $ with $ k_n(s)=h_n(s)/s^{1/2n}=n(s^{-1/2n}-s^{1/2n}). $ The derivative of the function $u\mapsto(\mathrm e^{cu/2}-\mathrm e^{-cu/2})/u$ is proportional to $\ell(u)$ with $ \ell(u)=cu(\mathrm e^{cu}+1)-\mathrm e^{cu}+1. $ Now, $\ell(0)=0$ and $\ell'(u)=c(1+cu\mathrm e^{cu})$, thus, for every $c\geqslant0$, $\ell'(u)\geqslant0$ for every $u\geqslant0$, and in particular $\ell(u)\geqslant0$, hence the function $u\mapsto(\mathrm e^{cu/2}-\mathrm e^{-cu/2})/u$ is nondecreasing on $u\geqslant0$.
Applying this to $u=1/n$ and $c=-\log(s)\geqslant0$ for every $s$ in $(0,1)$, one sees that the sequence $(k_n)_n$ is nonincreasing. Thus, the sequence $(g_n)_n$ is nondecreasing. The argument used for $(x_n)$ then shows that $(y_n)$ is nondecreasing.