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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?

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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.

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