# 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?

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.