Finding all positive real functions satisfying $xf(y)+f(f(y))\leq f(x+y)$ Find function $f: \mathbb{R}_{> 0}\rightarrow \mathbb{R}_{> 0}$ such that: $xf(y)+f(f(y))\leq f(x+y)$ for all positive $x$ and $y$?
That problem made me think a lot. This is the first time I solve functional inequality. Please show me the way to solve such problems. 
From the inequality, can we prove that $f$ is injective or surjective?
 A: $f(f(y))-f(x+y)\leq -xf(y)$.
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 $xf(y)+f(f(y))\leq f(x+y)$.
A: We first prove that $f(x)\leq x$. Suppose that for some $k$, $f(k)>k$. Then letting $x=f(k)-k$ and $y=k$ in the original equation shows that $$(f(k)-k)f(k)+f(f(k))\leq f(f(k)) \implies (f(k)-k)f(k)\leq 0$$ which is clearly a contradiction since we supposed $f(k)-k$ was positive and $f$ attains positive real values. Hence $f(x)\leq x$ for all $x\in\mathbb{R^+}$. 
Now fix a constant $c>0$. Note that for sufficiently large $N$, we have that $$\begin{align*} N(f(c)-1)&>c-f(f(c))\qquad  \text{ (true for sufficienty large $N$)} \\ \implies Nf(c)-N&>c-f(f(c)) \\ \implies Nf(c)+f(f(c))&>N+c\tag{1} \end{align*}$$
But if we put $x=N$ and $y=c$ into our original equation, we find that, on the contrary, $$Nf(c)+f(f(c))\leq f(N+c)\leq N+c$$ where the last inequality comes from the fact that $f(x)\leq x$. This clearly contradicts $(1)$, hence no such functions exist.
