Let $P(x,y)$ denote the assertion:

$f(f(x))> \frac{n}{n+1} x$ Now take $n\to +\infty ~$ to get $~f(f(x))\geq x$ Using this inequality in $P(x,y)$ gives us: $$xf(x)\geq yf(y)\implies xf(x)=c\implies f(x)=\frac{c}{x}~ \forall x\in \mathbb{R_+}$$Which is indeed a solution for any $c>0$.$~\blacksquare$

The answer is $f(x)=\frac cx$ for constant $c,$ which clearly works. First taking $x=y$ gives $f(f(x))\le x.$ Now rearrange the inequality to $xf(x)\ge f(y)(f(f(x))+y-x).$ This becomes \[\frac{xf(x)}{yf(y)}\ge1+\frac{f(f(x))-x}y.\]Now the RHS gets arbitrarily close to $1$ from below (or equals $1$) as $y$ gets large, for fixed $x.$ Additionally this gives us that as $y$ gets large, $f(y)$ gets arbitrarily close to $0.$ Then setting $x=f(y)$ in the original inequality gives $y-f(f(y))\le f(y)-f(f(f(y)))<f(y),$ so $y-f(f(y))$ gets arbitrarily close to $0$ for large $y.$ Now if we swap variable names, we get \[\frac{yf(y)}{xf(x)}\ge 1+\frac{f(f(y))-y}x.\]If we fix $x$ and make $y$ arbitrarily large, the RHS gets arbitrarily close to $1$ from below. Now we have arbitrarily tight bounds on $\frac{xf(x)}{yf(y)}$ in both directions. Thus fixing some positive reals $x,z$ gives us $\frac{xf(x)}{yf(y)}$ and $\frac{yf(y)}{zf(z)}$ are arbitrarily close to $1,$ so their product, which is constant, must be $1.$ Thus $xf(x)$ is some constant $c$ for all $x,$ and thus $f(x)=\frac cx.$

Solving this during TST was an extremely proud moment for me. Let $P(x,y)$ denote the assertion $x \cdot (f(x)+f(y)) \geq (f(f(x))+y) \cdot f(y)$. $P(x,x) \implies 2xf(x) \geq f(x)f(f(x))+xf(x) \implies x \geq f(f(x))$. $P(f(x),x) \implies f(f(x))+f(x) \geq f^3(x)+x \implies f(x)-f^3(x) \geq x-f(f(x))$. Now assume $f(f(a))>a$ for some $a \in \mathbb{R}^+$. Then put $a-f(f(a))=c$, note that for all positive integers $n$, we may inductively show that $f^{2n}(a)-f^{2n+2}(a) \geq c$. But this would imply that there exists some $n$ such that $f^n(a)<0$, contradiction. Hence $f(f(x))=x$ for all $x$. Putting this back into the original equation, we get $x \cdot (f(x)+f(y)) \geq (x+y) \cdot f(y) \implies xf(x) \geq yf(y)$ for all $x,y$. So $xf(x)=yf(y)$ for all $x,y$. Put $xf(x)=k$ for some constant $k$, we see that $f(x)=\frac{k}{x}$. It is easy to check that this is a solution. Hence the solutions are $f(x)=\frac{k}{x}$ for some constant $k \in \mathbb{R}^+$.

We take $P(f(x), x)$ to get $f(x)(f(f(x))+f(x)) \geq f(x)(f(f(f(x))) + x) \iff f(f(x)) + f(x) \geq f(f(f(x))) + x$ Thus $f(x) + f^2(x) \geq f^3(x) + x$. Take $x$ to $f(x) \implies f^2(x) + f^3(x) \geq f^4(x) + f(x)$. Adding them, we get $2 f^2(x) \geq f^4(x) + x$. Easy induction to get: $f^{2n}(x) \leq x + n(f^2(x) - x)$ If $f^2(x) < x \implies f^{2n}(x) < 0$, for $n$ large enough. So, $f^2(x) \geq x$. $P(x, x) \implies 2x f(x) \geq f(x) (x + f^2(x)) \iff 2x \leq x + f^2(x) \implies f(f(x)) = x$ $P(x, y) \implies x f(x) \leq y f(y)$. Swapping $x$ and $y$, we get $f(x) = \frac{c}{x}$, for some constant $c$. This function clearly works.

The only solutions are $f(x) = \frac{c}{x}$ for some positive real constant $c$. These clearly work. Now we prove they are the only solutions. Let $P(x,y)$ denote the given assertion. $P(x, x): 2xf(x) \ge (f(f(x)) + x) f(x) \implies f(f(x)) \le x$. $P(f(x), x): f(x) (f(f(x)) + f(x)) \ge (f^3(x) + x)f(x)$, so $f(f(x)) + f(x) \ge f^3(x) + x$, implying that $f(x) - f^3(x) \ge x - f(f(x))$. Claim: $f$ is an involution Proof: Consider some $x$ where $f(f(x)) < x$. Then let $d = x - f(f(x)) > 0$. We have $d \le f(x) - f^3(x) \le f(f(x)) - f^4(x) \ldots, $ so $d \le f^n(x) - f^{n+2}(x)$ for any nonnegative integer $n$, meaning that $f^{n+2}(x) \le f^n(x) - d$. Next we induct to show that \[f^{2n}(x) \le x - n\cdot d\]for any positive integer $n$. The base case $n = 1$ holds from $f(f(x)) = x - d$. Suppose it was true for $2k$. Then we have $f^{2k + 2} (x) \le f^{2k}(x) - d \le x - k \cdot d - d = x - (k+1)d$, as desired. Since $d > 0$, choose $n$ sufficiently large so that $x - n \cdot d < 0$. Then $f$ becomes negative, absurd. $\square$ Now the equation becomes $x(f(x) + f(y)) \ge (x + y) f(y)$, so $xf(x) \ge y f(y)$ and since $y f(y) \ge xf(x)$ also by swapping $x,y$, we have that $xf(x) = yf(y)$ for all reals $x,y$, so $xf(x) = f(1)$, meaning $f(x) = \frac{f(1)}{x}$, as desired.

As usual, $P(x,y)$ denotes the assertion of $(x,y)$ into the functional inequality, and $f^k(x)$ denotes an iteration rather than a power. First, $P(x,x)$ yields $x \geq f(f(x))$, and then $P(f(x),x)$ gives \[f(x)(f^2(x)+f(x)) \geq (f^3(x)+x)f(x)\Longrightarrow f^2(x) + f(x) \geq f^3(x) + x.\]Plugging $f(x)$ in the last inequality and summing: \begin{align*} f^3(x) + f^2(x) &\geq f^4(x) + f(x)\\ f^2(x) + f(x) &\geq f^3(x) + x\\ \Longrightarrow (f^3(x) + f^2(x)) + (f^2(x) + f(x)) &\geq (f^4(x) + f(x)) + (f^3(x) + x)\\ \Longrightarrow f^2(x) - f^4(x) &\geq x - f^2(x). \end{align*}Hence, if $a_k(x) = f^{2k}(x) - f^{2k+2}(x)$, we know from before that $a_0(x) \geq 0$, but if $a_0(x)>0$ for some $x$, then from the last inequality, the sequence $a_k(x)$ is monotonically increasing. However, this implies $f^{2k}(x) \leq x - k(x-f^2(x))$, which for large enough $k$ becomes negative, contradiction. Therefore, $f^2(x) = x$ for all $x>0$, and the original functional inequality becomes simply $xf(x) \geq yf(y)$. This is possible only if $xf(x)$ is constant, i.e. $f(x) = \frac{c}{x}$ for all $x$ and some positive constant $c$. These functions clearly work, so we're done.

let $$P(x,y):x(f(x)+f(y))\ge (ff(x)+y)f(y)$$ $$P(x,x):x\ge ff(x)$$ $$P(f(y),y):ff(y)+f(y)\ge fff(y)+y$$ $$P(ff(y),f(y)):fff(y)+ff(y)\ge ffff(y)+f(y)$$ So $$2ff(x)\ge y+ffff(y)$$ $$ff(y)-ffff(y)\ge y-ff(y)$$ if exist a number $y$ such that $$y>ff(y)$$ We can define $ C \colon \mathbb N_{>0} \to \mathbb R_{\ge 0}$ is a function such that $$f^{2i-2}(y)=f^{2i}(y)+C(1)+C(2)+.....+C(i-1)+C(i)$$ and$C(1)>0$ because $y=ff(y)+C(1)$ it appears that $f^{2i-2}(y)-C(1)\ge f^{2i}(y)$ and because $C(1)$ is a constant there is a positive integer $k$ such that $$f^{2k}(y)<0$$ wrong result So $C(1)=0$ and $y=ff(y)$ by editing in the main inequality $$xf(x)\ge yf(y)$$ and this lead to be $f(x)=\frac{c}{x}$. where $c$ is a positive real number and these functions work

The answer is $f(x) \equiv \tfrac{c}{x}$ where $c > 0$. Let $P(x,y)$ denote the given assertion. We have that $P(x,x)$ gives $f(f(x)) \le x $ and $P(f(x), x)$ gives $x - f(f(x)) \le f(x) - f(f(f(x)))$. This implies that \[x - f^2(x) \le f^2(x) - f^4(x) \le f^4(x) - f^6(x) \cdots\] Claim: $f(f(x)) = x$ for all $x$ Assume for contradiction that there exists an $x$ such that $x > f(f(x))$. Then the sequence $f^{2k}(x)$ will always skip down by at least $x-f(f(x))$ as $k$ increments. Therefore, $f^{2k}(x)$ will be negative for sufficiently large enough $k$, resulting in the desired contradiction. We can easily plug in the previous claim into the assertion to get $xf(x) \ge yf(y)$ and from swapping we obviously get $yf(y) \ge xf(x)$, implying $xf(x)$ is constant. Therefore we have $f(x) \equiv \tfrac{c}{x}$ which clearly works.

We claim the only solution is $f(x) = \frac cx$ for positive $c$ which is easy to see that these work. Let $P(x,y)$ denote the assertion. By taking $P(f(y),y)$, we get that \[y - f(f(y)) \le f(f(f(y))) - f(y)\]But this means \[y - f(f(y)) \le f^4(y) - f^2(y) \cdots\]So we we have $f(f(y)) = y$, otherwise repeat the above pattern until we get a $f^{2n}(y) < 0$ which is a contradiction. Plug this back in to get \[x(f(x) + f(y)) \ge (x+y)f(y) \implies xf(x) \ge yf(y) \implies xf(x) = yf(y) \: \forall x,y \in \mathbb{R}_{>0}\]So we recall our original solutions.

Easy problem $P(x,x): x \ge f(f(x))$ $P(f(y),y): f(y)-f(f(f(y))) \ge y-f(f(y))$ By repeat iteration, we get sequence ${f^{2k+2}(y)-f^{2k}(y)}$ does not decrease But when $y-f(f(y))>0$, this will lead to $f^{2N}(y)<0$ with sufficiently large N by summing up the differences So $y=f(f(y))$ then we can easily know $xf(x)=yf(y)$ for all positive real number $x,y$, and this lead to $f(x)=\frac{c}{x}$

Am I right?

First, take $y = x$ to get $x\ge f^2(x)$. Now fix a real $a_0 > 0$ and define the sequence $a_n = f^n(a_0)$ for each $n\ge 0$. We claim the following: Claim 1: $a_{n+1} - a_{n+3}\ge a_{n} - a_{n+2}$. proof: Take $(x,y) = (a_{n+1}, a_n)$ in the FE to get $a_{n+1}a_{n+2} + a_{n+1}^2\ge a_{n+3}a_{n+1}+ a_na_{n+1}$. Divide by $a_{n+1}$ and rearrange to finish $\square$. Claim 2: $a_{2n}\le a_0 - (a_0-a_2)n$ proof: Induct on $n$. Base case of $n = 0$ is obvious, now suppose it holds for some $n$. Then \[ a_{2n+2}\le 2a_{2n} - a_{2n-2} = a_{2n} -(a_{2n-2} - a_{2n}) \le a_{2n} - (a_0 - a_2)\le a_0 - (a_0-a_2)(n+1),\]done $\square$ A consequence of claim 2 is that $\{a_{2n}\}_{n\ge 0}$ is unbounded from below if $a_0-a_2 > 0$. But that's impossible since $\{a_{2n}\}_{n\ge 0}$ is a sequence of positive reals. Since $a_0 - a_2\ge 0$, we are then forced to have $a_0 - a_2 = 0$. Hence $a_0 = a_2$, meaning $f(f(a_0)) = a_0$ for all $a_0$. The original FE then rewrites as \[ xf(x) + xf(y)\ge xf(y) + yf(y)\implies xf(x)\ge yf(y)\]for all $x,y > 0$. Swap $x,y$ to get $xf(x) = yf(y)$ for all $x,y$. Take $y = 1$ to finally get $\boxed{f(x) = f(1)/x}$ for all $x$, which clearly works.

Given condition is equivalent to: $P(x,y):\ \left(f\left(f\left(x\right)\right)+y-x\right)f\left(y\right)\le xf\left(x\right)$ $P(x,x):\ f(f(x))\leq x$. Claim: $f(f(x))=x$ for every $x>0$. Proof. $P(f(x),x):\ 0\leq x-f(f(x)) \leq f(x) - f(f(f(x)))$ Hence, if $t=x-f(f(x))>0$, then $f^n(x)-f^{n+2}(x)\geq t$ for every $n\geq 0$. So, for any $n$, we have \[f^{2n}(x)\geq t\Longrightarrow f^{2n-2}(x)\geq 2t, f^{2n-4}(t)\geq 3t, \ldots, x\geq nt,\]which is impossible for sufficiently large $n$. This forces $t=0$, and $f(f(x))=x$ for every $x>0$. $\blacksquare$ This means $P(x,y)\Longleftrightarrow xf(x)\geq yf(y)$, while $P(y,x)\Longleftrightarrow yf(y)\geq xf(x)\Longrightarrow xf(x)=yf(y)$. So, $xf(x)=c$ for some constant $c>0$. So, $f(x)=\frac cx$ for all $x$, which indeed satisfies the original condition. $\blacksquare$

The only solution is $f(x)=\frac{c}{x}, c\in \mathbb{R}_{>0}$, which turns the inequality into an equality. Let $P(x,y)$ be the given assertion. Checking $P(x,f(x))$ gives that $x\geq f(f(x))$. Define $d(x)=x-f(f(x))$. The assertion $P(f(x),x)$ gives that $d(f(x))\geq d(x)$. This implies that $d(f(f(x)))\geq d(x)$. However we have that $$(k+1)d(x)\leq d(x)+d(f^2(x))+\dots+d(f^{2k}(x))=x-f^{2k+2}(x)\leq x$$By taking $k\rightarrow \infty$ we must have that $d(x)=0$, that is $f(f(x))=x$. This simplifies the assertion to $xf(x)\geq yf(y)$. Then however both sides must be constant, implying the above solution.

shanelin-sigma wrote: Am I right? Why does $min(G)$ exist?I think you should use the supremum and infimum principle instead of directly setting the minimum value

Yue-Zhang_3906 wrote: Why does $min(G)$ exist?I think you should use the supremum and infimum principle instead of directly setting the minimum value Oh no, I made an elementary mistake. Sorry but how could I fix it?

Based on my solution during my country's TST which it appeared in which also seems to be the first proof of this sort on this thread. Denote the asserion by $P\left(x,y\right)$. From $P\left(x,x\right)$ we have $f\left(f\left(x\right)\right)\leq x$. Now take $x\to0$ in $P\left(x,y\right)$ and denote $c_1=\liminf_{x\to0}{xf\left(x\right)}$: $$0<f\left(f\left(x\right)\right)\leq x\to0\Rightarrow yf\left(y\right)\leq c_1$$So we have a function that is at most it's liminf meaning $\lim_{x\to0}{xf\left(x\right)}=c_1$. Now $f\left(y\right)\leq\frac{c_1}{y}$ meaning $\lim_{y\to\infty}{f\left(y\right)}=0$ so taking $y\to\infty$ in $P\left(x,y\right)$ and denoting $c_2=\limsup_{y\to\infty}{yf\left(y\right)}$ gives $xf\left(x\right)\geq c_2$ and similarly to earlier, the function is always at least it's limsup meaning $\lim_{x\to\infty}{xf\left(x\right)}=c_2$. So for now we have: $$c_1\geq xf\left(x\right)\geq c_2$$Now notice $c_1\leq c_2$ which follows from taking $x\to\infty$ in $f\left(x\right)f\left(f\left(x\right)\right)\leq xf\left(x\right)$ so we have $xf\left(x\right)$ is constant i.e. $f\left(x\right)=\frac{c}{x},c\in\mathbb{R}_{>0}\thinspace\forall x\in\mathbb{R}_{>0}$ which fits.

Firstly, we rewrite the given inequality as $xf(x)\geqslant f(y)(f(f(x))-x+y)$. Notice that $\lim_{x \to \infty}f(x)=0$. To see this, fix $x$ and take $y$ to infinity. The left hand side of the inequality is constant, but the right hand side is asymptotically $yf(y)$, so $f(y)$ must go to 0. Again, by taking $y$ to infinity we have \[xf(x)\geqslant \limsup_{y \to \infty}yf(y)\implies M:=\inf \{xf(x) | x \in \mathbb{R}_{>0}\}\geqslant \limsup_{y \to \infty}yf(y).\]Plugging in $x=y$ gives $x\geqslant f(f(x))$ for all $x$. Multiplying this by $f(x)$ gives $xf(x)\geqslant f(x)f(f(x))$ and hence \[M \geqslant \limsup_{x\to \infty}f(x)f(f(x)).\]Setting $x$ to $f(x)$ in the initial inequality and taking $x$ to infinity gives \[M \geqslant \limsup_{x\to \infty}f(x)f(f(x)) \geqslant yf(y) -f(y)\limsup_{x\to \infty}(f(x)-f(f(f(x)))),\]but since $\lim_{x \to \infty}f(x)=0$ and $f(x)\geqslant f(f(f(x)))$, we get that the last limsup is equal to 0. Finally we get that \[M\geqslant yf(y)\]for any $y$. By the definition of $M$, we must have an equality for all $y$, so $f(x)=\frac{M}{x}$ for all $x$. This is in fact a solution for any positive $M$.

Straightforward enough.

\begin{align*} s&\geq f(f(s))\tag{$P(s,s)$}\\ &\geq\frac{\alpha f(\alpha)}{f(s)}-f(\alpha)\tag{$P(f(s),\alpha)$ dropping the $f(f(x))$ term}\\ &\geq\frac{\alpha f(\alpha)}{\beta f(\beta)}(s-\beta)-f(\alpha)\tag{$P(\beta,s)$ dropping the $f(f(x))$ term}\\ \end{align*}which fails for $\alpha f(\alpha)>\beta f(\beta)$ and sufficiently large $s>\beta$. We conclude.

based problem

We uploaded our solution https://calimath.org/pdf/ISL2023-A4.pdf on youtube https://youtu.be/aDr1Ai0uwgw.

We rewrite as $f(y) (x-f^2(x)) \ge yf(y) - xf(x)$. Putting $(x,x)$ gives $x\ge f^2(x) \Rightarrow f^{n} (x) \ge f^{n+2} (x)$. Notice that $$x\ge f^2(x) \ge f^4(x) \ge \ldots > 0.$$This implies that $\lim\limits_{n\to \infty} f^{2n} (x) - f^{2n + 2}(x) = 0$. Putting $(f^{2n}(x),y)$ gives us $$f(y)(f^{2n}(x)-f^{2n+2}(x))\ge yf(y) - f^{2n}(x)f^{2n+1}(x) \ge yf(y) - xf(x).$$Fixing $x,y$ and taking $n\to\infty$ implies that $0 \ge yf(y)-xf(x)$, thus $xf(x) \equiv c$ for some $c > 0$.

Let $P(x, y)$ be the assertion in the question. \[P(x, x)\]\[x\leq f(f(x))\]Rearrange the inequality to: \[\frac{xf(x)}{yf(y)}\geq \frac{f(f(x))-x}{y}+1\]Thus we get that $\frac{xf(x)}{yf(y)}$ gets abitrarily close to $1$ from below as we make $y$ large. By swapping variables we get that $\frac{yf(y)}{xf(x)}$ gets abitrarily close to $1$ from below as we make $x$ large. So we get that $xf(x)=c$ for some fixed $c$. Thus the only solutions are $f(x)=\frac{c}{x}$.

it is very easy such that we just try monotonnia of function F ,and get $f(x)=\frac{c}{x}$

The answer is $f(x) = \frac cx$ for every positive real $c$, which work. Setting $y = f(x)$ yields \[x(f(x)+f(f(x)) \geq f(f(x))(f(x)+f(f(x)))\]for every $x$, i.e. $x \geq f(f(x))$ for each $x$. I will show that this must actually be an equality: Claim: $f(f(x)) = x$ for every real number $x$. Proof: Assume for the sake of contradiction that there exists an $\varepsilon > 0$ such that $f(f(x_0)) < x_0 - \varepsilon$ for some fixed $x_0$. Setting $f(x_0), x_0$ in the original equation, \[f(x_0)(f(f(x_0)) + f(x_0)) \geq f(x_0)(f(f(f(x_0))) + x_0)\]so in particular \[f(x_0) - f(f(f(x_0))) \geq f(f(x_0)) - f(x_0) > \varepsilon.\]Repeatedly applying this argument, it follows that \[f^{2n-2}(x_0) - f^{2n}(x_0) < \varepsilon\]for each $\varepsilon$. But picking an $n$ such that $n\varepsilon > x_0$, $f^{2n+2}(x_0) < 0$, which is a contradiction. $\blacksquare$ Thus we have \[x(f(x) + f(y)) \geq (x+y)f(y) \iff xf(x) \geq yf(y)\]for every pair of real numbers $(x, y)$. This implies $xf(x) = yf(y)$ for all $(x, y)$, so $xf(x) = c$, and $f(x) = \frac cx$ for some $c > 0$ all work.