Let $P(x,y)$ be the assertion. P(1,1): $f(f(1))\leq 1$ so $f(f(1))=1$ P(1,f(1)):$2f(1)\leq 2$, so $f(1)=1)$ P(1,y):$f(y)-y\geq f(1)-1=0$, so $f(y)\geq y$ P(x,x):$2x\leq x+f(x)\leq f(x)+f(f(x))\leq 2x$, thus we have $f(x)=x$ for all x. Let's check it: $x+yx\leq x+xy$, thus the only function is $f(x)=x$

If you plug in x=y=1, you get f(1)+f(f(1))<=1+f(1) and f(f(1))<=1 so f(f(1))=1. Now, we try to plug x=1 and y=1 respectively(not in once), and we get something like f(1)+y<=1+f(y) and f(x)+f(f(x))<=x+xf(1). From the first equation it follows f(y)-y>=f(1)-1. Since we know that f maps natural numbers to natural numbers, and we know that that equation works for all natural numbers y, we can use our prevoious observation and plug y=f(1), so we get something like f(f(1))-f(1)=1-f(1)>=f(1)-1 so f(1)<=1 and f(1) must be 1. Ok, so now we look back again to previous equations and from f(1)+y<=1+f(y) it follows y<=f(y) and from the f(x)+f(f(x))<=x+xf(1)=x+x=2x<=2f(x) so that means f(f(x))<=f(x), but since x<=f(x), we get also for x=f(x): f(x)<=f(f(x)), combining these two things we get f(x)=f(f(x)). Now from x=x and y=1 we know f(x)+f(f(x))=2f(x)<=2x so f(x)<=x and hence f(x)=x is the only possibility. It's easy to check. Hope that helps

That's immensely easy problem for a senior category

Well good for you I guess?

BartSimpsons wrote: Find all functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that the inequality $$f(x)+yf(f(x))\le x(1+f(y))$$holds for all positive integers $x, y$. Proposed by Adrian Beker. x,y--->1 f(f(1))=1.Of course, if we take x,y--->f(1), then it implies f(1)^2-2f(1)+1<=0.So f(1)=1. y--->1 f(x)+f(f(x))<=2x and y--->f(x) yields f(x)<=x.But x--->1 gives y<=f(y).Thus f(x)=x for all real numbers x. The problem was probably difficult for you Bye the way,Welcome to Aops!

JANMATH111 wrote: If you plug in $x=y=1$, you get $f(1)+f(f(1))<=1+f(1)$ and $f(f(1))<=1$ so $f(f(1))=1$. Now, we try to plug $x=1$ and $y=1$ respectively (not in once), and we get something like $f(1)+y<=1+f(y)$ and $f(x)+f(f(x))<=x+xf(1)$. From the first equation it follows $f(y)-y>=f(1)-1$. Since we know that $f$ maps natural numbers to natural numbers, and we know that that equation works for all natural numbers $y$, we can use our previous observation and plug $y=f(1)$, so we get something like $f(f(1))-f(1)=1-f(1)>=f(1)-1$ so $f(1)<=1$ and $f(1)$ must be $1$. Ok, so now we look back again to previous equations and from $f(1)+y<=1+f(y)$ it follows $y<=f(y)$ and from the $f(x)+f(f(x))<=x+xf(1)=x+x=2x<=2f(x)$ so that means $f(f(x))<=f(x)$, but since $x<=f(x)$, we get also for $x=f(x):$ $f(x)<=f(f(x))$, combining these two things we get $f(x)=f(f(x))$. Now from $x=x$ and $y=1$ we know $f(x)+f(f(x))=2f(x)<=2x$ so $f(x)<=x$ and hence $f(x)=x$ is the only possibility. It's easy to check. Hope that helps EDIT: LaTeX

No, the problem wasn't difficult for me, but it was an average difficulty for problem 1 in emc

Here is a more general question: https://artofproblemsolving.com/community/c6h1565528

Let $P(x,y)$ be the assertion of $x$ and $y$ to the given functional inequality. We will prove that only $\boxed{f(x) = x}$ works. It is easy to check that this function holds. Claim 01. $f(1) = 1$. Proof. First, $P(1,1)$ gives us $f(f(1)) \le 1$ which forces $f(f(1)) = 1$ since $f:\mathbb{N} \to \mathbb{N}$. Now, $P(f(1), f(1))$ gives us $1 + f(1)^2 \le 2f(1)$, which forces $f(1) = 1$. Claim 02. $f(x) \ge x$ for all $x \in \mathbb{N}$ Proof. $P(1,x)$ gives us $1 + x \le 1 + f(x)$ for all $x \in \mathbb{N}$. Claim 03. $f(x) \le x$ for all $x \in \mathbb{N}$. Proof. By Claim 02, \[ f(x) + yf(x) \le f(x) + yf(f(x)) \le x(1 + f(y)) \Rightarrow \frac{f(x)}{x} \le \frac{f(y) + 1}{y + 1} \]Let $y = 1$, we then have $f(x) \le x$ for all $x \in \mathbb{N}$. Therefore, we are done as we conclude $f(x) = x$ for all $x \in \mathbb{N}$.

Nice.

BartSimpsons wrote: Find all functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that the inequality $$f(x)+yf(f(x))\le x(1+f(y))$$holds for all positive integers $x, y$. Proposed by Adrian Beker. Let \(P(x,y)\) denote the assertion of the given functional equation. \(P(1,1)\) gives us that \(f(f(1))\leq1\), so \(f(f(1))=1\). Next, \(P(1,y)\) gives us that \[f(y)-y\geq f(1)-1\geq0\]so \(f(x)\geq x\) for all \(x\). Finally, we have \[xf(y)+x\geq yf(f(x))+f(x)\geq yf(x)+x\]so \(xf(y)\geq yf(x)\) for all \(x,y\), interchanging \(x\) and \(y\) gives us that \(xf(y)=yf(x)\) or \(f(x)=cx\) for all \(x\) where \(c\) is a fixed constant. Plugging in gives us that \[(c-1)(cxy+x)\leq 0\]and as \(cxy+x\geq0\), we must have \(c=1\), implying that \(f(x)=x\) for all \(x\in\mathbb{N}\).

Let $P(x,y)$ denote the assertion $f(x)+yf(f(x))\le x+xf(y)$. $P(1,1)\Rightarrow f(f(1))=1$ $P(1,f(1))\Rightarrow f(1)=1$ $P(1,x)\Rightarrow f(x)\ge x$ $P(x,x)\Rightarrow x+xf(x)\ge f(x)+xf(f(x))\ge x+xf(f(x))\ge x+xf(x)$, so equality holds and $\boxed{f(x)=x}$ for all $x$.

Why so short? Denote what $P(x,y)$ is. $P(1,1)$ and $P(1,f(1))$ give $f(1)=1.$ Then $P(1,x)$ gives $f(x)\geq x.$ Now it is not hard to get $f(x)\leq x.$ So $f(x)=x.$

We claim $f(x)=x$ is the only solution. $P(1, 1) \rightarrow f(f(1)) \leq 1 \rightarrow f(f(1)) = 1$ $P(1, f(1)) \rightarrow f(1) \leq 1 \rightarrow f(1) = 1$ $P(1, x) \rightarrow f(x) \geq x$ $P(x, x)$ combined with $f(x) \geq x$ gives $f(x) \leq x$. Thus, $f(x) = x$. $\blacksquare$

Let $P(x,y)$ denote the given assertion. $P(1,1): f(1)+f(f(1))\le f(1)+1\implies f(f(1))\le 1\implies f(f(1))=1$. $P(1,f(1)): 2f(1)\le 2\implies f(1)\le 1\implies f(1)=1$. $P(1,x): x+1\le f(x)+1\implies f(x)\ge x$. $P(x,x): f(x)+xf(f(x))\le xf(x)+x$. Note that $f(x)+xf(f(x))\ge x+xf(x)$, so they must be equal. However, if $f(x)>x$, then $f(x)+xf(f(x))>x+xf(x)$, a contradiction. So $\boxed{f(x)=x}$ for all $x$, which works.

After $P(1,1)$, $P(1,f(1))$ and $P(1,x)$ you can also finish with $P(x,1)$, which feels more natural for me than $P(x,x)$.

Let $p(x, y)$ denote the given assertion. $$p(1, 1) \implies f(1)+f(f(1)) \le 1+f(1) \implies f(f(1)) \le 1 \implies f(f(1))=1.$$We know that because the functions takes in and out natural numbers. Let $f(1)=u,$ we have $$p(1, u) \implies f(1)+uf(f(1)) \le 1+f(u) \implies f(1)+u \le 1+u \implies f(1) \le 1 \implies f(1)=1.$$Now, letting $x=1, y=x,$ we have $$p(1, x) \implies f(1)+xf(f(1)) \le 1+f(x) \implies x+1 \le f(x)+1 \implies f(x) \ge x.$$We also have $$p(x, 1) \implies f(x)+f(f(x)) \le x(1+f(1)) \implies f(x)+f(f(x)) \le 2x,$$using $f(x) \ge x$ we have $$f(x)+f(f(x)) \ge x+f(x) \implies f(x)+x \le 2x \implies f(x) \le x.$$But we have $f(x) \le x, f(x) \ge x,$ therefore the only function that satisfies the given functional equation is $\boxed{f(x)=x}.$

Let $ P(x,y) $ denote the given assertion. $ P(1,1) \Rightarrow f(f(1)) \le 1 \Rightarrow f(f(1)) = 1 $. $ P(f(1),f(1)) \Rightarrow 1 + f(1)^2 \le 2f(1) \Rightarrow (f(1) - 1)^2 \le 0 \Rightarrow (f(1) - 1)^2 = 0 \Rightarrow f(1) = 1 $. $ P(1,x) \Rightarrow x \le f(x) $.(*) $ P(1,f(x)) \Rightarrow f(x) \le f(f(x)) \Rightarrow $ by (*) $ x $ $\le$ $ f(f(x)) $.(**) $ P(x,1) \Rightarrow f(x) + f(f(x)) \le 2x $. By (**) $ f(x) \le x $.(***) By (*) and (***) $ \longrightarrow f(x) = x $ for all $ x \in \mathbb{N} $.

Note that $x = 1 \implies f(y) \ge y$. Now, we have $$x + yf(x) \le f(x) + yf(f(x)) \le x(1+f(y)) = x + xf(y)$$$$\implies yf(x) \le xf(y).$$From symmetry, $yf(x) \ge xf(y)$. So $xf(y) = yf(x)$, i. e. $f(x) = cx$ for some $c \in \mathbb{N}$. It may be verified that $c = 1$ is the only working solution (by putting $x = y = 1$, for example), so the only solution is $\boxed{f \equiv id}$. $\square$

Solution without finding $f(1) or f(f(1))$. $P(x,y) \implies f(x) \geq x$. $x(1+f(f(y))) \geq yf(f(x))+f(x) \geq yf(f(x))+x \implies xf(y) \geq yf(f(x))$. By taking $y=f(x)$ $\implies$ $x \geq f(x)$.