Finding all positive real functions satisfying $xf(y)+f(f(y))\leq f(x+y)$

$f(f(y))-f(x+y)\leq -xf(y)$.

As $xf(y)>0$, $f(f(y))-f(x+y)<0$.

Now, if $f(y_0)>y_0$ for some $y_0>0$, then we can put $x=f(y_0)-y_0$ and $y=y_0$ which yields $f(f(y_0))-f(f(y_0))<0$ which is clearly wrong.

Therefore, $$f(y)\leq y$$

Now, as $f(f(y))>0$, $xf(y)<f(x+y)\leq x+y$

$x(f(y)-1)<y$ $\forall$ $x,y>0$.

If $f(y)>1$ for any finite $y>0$, by making $x$ arbitrarily large, we obtain a contradiction.

Thus, $$f(y)\leq 1$$

As $f(f(y))>0$, $f(x+y)\leq 1$ and $xf(y)+f(f(y))\leq f(x+y)$, $$xf(y)<1$$

As $f(y)>0$, by making $x$ arbitrarily large, we obtain another contradiction.

Therefore, there exists no function $f:\mathbb{R}_{>0}\to \mathbb{R}_{>0}$ such that $xf(y)+f(f(y))\leq f(x+y)$.