Functional inequality $f(y)+xf(x)≤yf(x)+f(f(x))$ Let $f:\mathbb{R}\to\mathbb{R}$ be a function sucht that:
$$
f(y)+xf(x)≤yf(x)+f(f(x))
$$
for all $x,y\in\mathbb{R}$. 
Show that 
$$
f(x)+yf(x+y)≤0
$$
for all $x,y\in\mathbb{R}$.
I tried some substitutions but nothing worth to mention came out. How to solve it?
 A: Let $P(x,y)$ be the assertion $f(y)+xf(x)≤yf(x)+f(f(x))$.
$$
P(0,f(y)):\space f(f(y))≤f(y)f(0)+f(f(0))
$$
Thus:
$$
P(x,y):\space f(y)+xf(x)≤yf(x)+f(f(x))≤yf(x)+f(x)f(0)+f(f(0))\implies \\
f(x)(x-y-f(0))≤-f(y)+f(f(0))\implies \\
f(x+y+f(0))x≤-f(y)+f(f(0))\implies \\
f(x+y)x≤-f(y-f(0))+f(f(0))\space\space (1)
$$
Therefore, if we set $x≤0$:
$$
P(y,x+y):\space f(x+y)+yf(y)≤(x+y)f(y)+f(f(y))\implies \\
f(x+y)x≥x^2f(y)+xf(f(y))\space\space (2)
$$
Combining $(1)$ and $(2)$ yields:
$$
x^2f(y)+xf(f(y))≤-f(y-f(0))+f(f(0))\space \forall x≤0\forall y\in\mathbb{R}\space\space (3)
$$
Suppose there exists a $z\in\mathbb{R}$ with $f(z)>0$. But if $x$ tends to $-\infty$ in $(3)$, we get a contradiction. Thus $f(y)≤0\space\forall y\in\mathbb{R}$. If we substitute $y=2f(0)$ in $(1)$ we get:
$$
f(x+2f(0))x≤0
$$
Thus, if $x<0$, we have $f(x+2f(0))=0$, i.e. $f(2f(0)-1)=0$.
$$
P(2f(0)-1,2f(0)-1):\space 0≤f(0)\implies f(0)=0
$$
Thus we conclude that $f(x)=0\space\forall x≤0\implies f(f(x))=0 \space\forall x≤0$. Thus we get:
$$
P(x+y,x): \space f(x)+(x+y)f(x+y)≤xf(x+y)\implies f(x)+yf(x+y)≤0
$$
And were done.
