Answer: All functions $f(x)=\frac{c}{x}$ for some $c>0$.

Solution 1. Let $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ be a function that satisfies the inequality of the problem statement. We will write $f^{k}(x)=f(f(\cdots f(x) \cdots))$ for the composition of $f$ with itself $k$ times, with the convention that $f^{0}(x)=x$. Substituting $y=x$ gives
$$x \geqslant f^{2}(x)$$
Substituting $x=f(y)$ instead leads to $f(y)+f^{2}(y) \geqslant y+f^{3}(y)$, or equivalently
$$f(y)-f^{3}(y) \geqslant y-f^{2}(y)$$
We can generalise this inequality. If we replace $y$ by $f^{n-1}(y)$ in the above inequality, we get
$$f^{n}(y)-f^{n+2}(y) \geqslant f^{n-1}(y)-f^{n+1}(y)$$
for every $y \in \mathbb{R}_{>0}$ and for every integer $n \geqslant 1$. In particular, $f^{n}(y)-f^{n+2}(y) \geqslant y-f^{2}(y) \geqslant 0$ for every $n \geqslant 1$. Hereafter consider even integers $n=2 m$. Observe that
$$y-f^{2 m}(y)=\sum_{i=0}^{m-1}\left(f^{2 i}(y)-f^{2 i+2}(y)\right) \geqslant m\left(y-f^{2}(y)\right) .$$
Since $f$ takes positive values, it holds that $y-f^{2 m}(y)<y$ for every $m \geqslant 1$. So, we have proved that $y>m\left(y-f^{2}(y)\right)$ for every $y \in \mathbb{R}_{>0}$ and every $m \geqslant 1$. Since $y-f^{2}(y) \geqslant 0$, this holds if only if
$$f^{2}(y)=y$$
for every $y \in \mathbb{R}_{>0}$. The original inequality becomes
$$x f(x) \geqslant y f(y)$$
for every $x, y \in \mathbb{R}_{>0}$. Hence, $x f(x)$ is constant. We conclude that $f(x)=c / x$ for some $c>0$.
We now check that all the functions of the form $f(x)=c / x$ are indeed solutions of the original problem. First, note that all these functions satisfy $f(f(x))=c /(c / x)=x$. So it's sufficient to check that $x f(x) \geqslant y f(y)$, which is true since $c \geqslant c$.

Let $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ be a function that satisfies the inequality of the problem statement. As in Solution 1, we prove that
$$f^{n}(y) \geqslant f^{n+2}(y)$$
for every $y \in \mathbb{R}_{>0}$ and every $n \geqslant 0$. Since $f$ takes positive values, this implies that
$$y f(y) \geqslant f(y) f^{2}(y) \geqslant f^{2}(y) f^{3}(y) \geqslant \cdots$$
In other words, $y f(y) \geqslant f^{n}(y) f^{n+1}(y)$ for every $y \in \mathbb{R}_{>0}$ and every $n \geqslant 1$.
We replace $x$ by $f^{n}(x)$ in the original inequality and get
$$f^{n}(x)-f^{n+2}(x) \geqslant \frac{y f(y)-f^{n}(x) f^{n+1}(x)}{f(y)}$$
Using that $x f(x) \geqslant f^{n}(x) f^{n+1}(x)$, we obtain
$$f^{n}(x)-f^{n+2}(x) \geqslant \frac{y f(y)-x f(x)}{f(y)}$$
for every $n \geqslant 0$. The same trick as in Solution 1 gives
$$x>x-f^{2 m}(x)=\sum_{i=0}^{m-1}\left(f^{2 i}(x)-f^{2 i+2}(x)\right) \geqslant m \cdot \frac{y f(y)-x f(x)}{f(y)}$$
for every $x, y \in \mathbb{R}_{>0}$ and every $m \geqslant 1$. Possibly permuting $x$ and $y$, we may assume that $y f(y)-x f(x) \geqslant 0$ then the above inequality implies $x f(x)=y f(y)$. We conclude as in Solution 1.