NikitosKh wrote: Determine, whether there exist a function $f$ defined on the set of all positive real numbers and taking positive values such that $f(x+y)\geqslant yf(x)+f(f(x))$ for all positive x and y? Let $P(x,y)$ be the assertion $f(x+y)\ge yf(x)+f(f(x))$ Let $u=1+\frac 1{f(1)}$ : $P(1,\frac 1{f(1)})$ $\implies$ $f(u)>1$ If $f(x)>x$ for some $x$, then $P(x,f(x)-x)$ $\implies$ $0\ge f(x)(f(x)-x)>0$, impossible So $f(x)\le x$ $\forall x>0$ and also $f(f(x))\le f(x)\le x$ $\forall x>0$ Let then $y>\frac{u-f(f(u))}{f(u)-1}$ : This is $yf(u)+f(f(u))>y+u$ and so $yf(u)+f(f(u))>f(y+u)$ in contradiction with $P(u,y)$ And so $\boxed{\text{No such function}}$

I am not sure, please tell if it is wrong (Almost sure it will be wrong ) $f(x+y)\ge yf(x)+f(f(x))$ Letting $$u=f(x), y=0 \implies u \ge f(u) \forall u=f(x) \in N$$Letting $$y=f(x)-x=u-x \implies 0 \ge u(u-x) \implies u\ge x \implies f(x) \ge x \forall x \in N$$So, $f(x)=x$ but it gives, $x+y \ge x+xy \implies 1 \ge x$ but $x \in N$ therefore $f(x)$ not equal to $x$ $\text{No such function}$

lian_the_noob12 wrote: I am not sure, please tell if it is wrong (Almost sure it will be wrong ) $f(x+y)\ge yf(x)+f(f(x))$ Letting $$u=f(x), y=0 \implies u \ge f(u) \forall u=f(x) \in N$$ You can not set $y=0$ since inequation is valid only for $x,y>0$

Suppose $f(x)-x>0$ for some $x$ then set $y=f(x)-x$ to get $0\geq (f(x)-x)f(x)$ which is a contradiction. So $f(x)\leq x$ for all $x$. So $x+y\geq yf(x)+f(f(x))$, which we implies that $x>y(f(x)-1)$, if $f(x)>1$ for some $x$ then this gives a contradiction. So $f(x)\leq 1$ for all $x$, so $1\geq yf(x)+f(f(x))>yf(x)$ for all $y$, taking appropriate value of $y$ this also gives contradiction. Thus $f$ does not exist.