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#1 h Apr 21, 2025, 12:59 PM

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Find all functions $f: \mathbb{R^+} \rightarrow \mathbb{R^+}$ such that:
$f\left(f(xy)+1\right)=xf\left(x+f(y)\right)$

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#2 h Apr 21, 2025, 1:40 PM

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Try $P(x,1)$ and isolating $x$ on the $RHS$ .
$\boxed{f(x)=\dfrac{1}{x}}$ works.

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#3 h Apr 23, 2025, 2:02 PM • 2 Y

Y by Alex-131, KevinYang2.71

The answer is $f(x) = \frac ax$ for any positive real $a$ , which clearly works. Now we show it's the only solution. Let $P(x,y)$ be the given assertion.

Claim 1: $f$ is surjective.
Proof: Fix $c$ in the image of $f$ and let $f(a) = c$ for some $a$ . Setting $y = \frac ax$ gives that $\frac{f(f(xy) + 1)}{x} = \frac{f(c+1)}{x}$ is in the image of $f$ for all positive reals $x$ , which takes on all positive reals. $\square$

$P(x,1): f(f(x) + 1) = xf(x + f(1))$ .

So by comparing $P(xy,1)$ and $P(x,y)$ , we have $xy f(xy + f(1)) = x f(x + f(y)) \implies y f(xy + f(1)) = f(x + f(y))$ Let $Q(x,y)$ be this assertion.

For $y \ne 1$ , $\frac{f(y) - f(1)}{y - 1}$ is not positive (otherwise we could set $x$ equal to it and get $y = 1$ from $Q(x,y)$ ), so if $f(y) > f(1)$ , then $y < 1$ , and if $f(y) < f(1)$ , then $y > 1$ . So if $y > 1$ , then $f(y) \le f(1)$ and if $y < 1$ , $f(y) \ge f(1)$ .

Thus, we have $xf(x + f(y)) = f(f(xy) + 1) \le f(1)$ .

Since $f$ is surjective, means that $f(x+y) \le \frac{f(1)}{x}$ for all positive reals $x,y$ .

Claim 2: $f(x) \le \frac{f(1)}{x}$ for all $x$ .
Proof: Suppose $f(c) > \frac{f(1)}{c}$ for some $c$ . Setting $y = c - x$ gives that $f(c) \le \frac{f(1)}{x}$ for $x < c$ . Setting $x$ arbitrarily close to $c$ gives a contradiction. $\square$

Claim 3: $f(f(x)) \ge x$ for all $x$ .
Proof: Suppose for some $c$ we had $f(f(c)) < c$ .

$Q(f(c), c - f(f(c))): f(c) f((c - f(f(c)) ) f(c) + f(1)) = f(c)$ , so $f((c - f(f(c)) ) f(c) + f(1)) = 1$ However, for any $k > f(1)$ , we have $f(k) \le \frac{f(1)}{k} < 1$ , a contradiction by setting $k = (c - f(f(c)) ) f(c) + f(1)$ . $\square$

Now, we have by setting $x = f(1)$ in claim 3 that $f(f(1)) \le 1$ , so combined with $f(f(1)) \ge 1$ , we have $f(f(1)) = 1$ . Now write $a = f(1)$ . We have $y f(xy + a) = f(x + f(y)),$ $f(a) = 1$ , and also $f(x) \le \frac ax$ for all $x \in \mathbb R^{+}$ .

Claim: $f(y) < a$ for all positive reals $y > 1$
Proof: We already know that $f(y) \le a$ . If $f(y) = a$ , then $f(f(y)) = 1 < y$ , absurd by claim 3. $\square$

For $y > 1$ , $Q(a - f(y), y): y f((a - f(y)) y + a) = 1$ , so $f((a - f(y)) y + a) = \frac 1y$ This means $\frac 1y \le \frac{a}{(a - f(y)) y + a} \implies y \ge y \left( 1 - \frac{f(y)}{a} \right) + 1$ Therefore, $y \cdot \frac{f(y)}{a} \ge 1$ , so $yf(y) \ge a \implies f(y) \ge \frac ay$ , but since we already have $f(y)\le \frac ay$ , we have $f(y) = \frac ay$ for each $y > 1$ , so $f(y) = \frac ay \forall y \ge 1$ .

Now, in $Q(x,y)$ , fix $y$ to be any positive real number and $x$ large enough so that $xy + a > 1$ .

We have $\frac{ay}{xy + a} = \frac{a}{x + f(y)}$ , so $\frac{ay}{xy + a} = \frac{ay}{(x+f(y)) y} \implies xy + a = xy + y f(y),$ so $y f(y) = a \implies f(y) = \frac ay$ for all positive reals $y$ .

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#4 h Apr 25, 2025, 1:23 AM • 3 Y

Y by megarnie, Alex-131, FPilipovic

The answer is $f(x)=\frac{a}{x}$ for a constant $a\in\mathbb{R}^+$ , which works.

Let $P(x,\,y)$ denote the assertion. From $P(x,\,1/x)$ we get $f(f(1)+1)=xf(x+f(1/x))$ so $f(x+f(1/x))=\frac{f(f(1)+1)}{x}$ . Thus $f$ is surjective.

Claim. $f$ is non-increasing.
Proof. Comparing $P(xy,\,z)$ and $P(xz,\,y)$ yields $xyf(xy+f(z))=f(f(xyz)+1)=xzf(xz+f(y))$ so $yf(xy+f(z))=zf(xz+f(y))$ . If $\frac{f(y)-f(z)}{y-z}>0$ , then $x:=\frac{f(y)-f(z)}{y-z}$ satisfies $xy+f(z)=xz+f(y)$ so $y=z$ , a contradiction. Thus $f(y)\leq f(z)$ for all $y>z>0$ . $\square$

For each $c\in\mathbb{R}^+$ , if $\lim_{x\to c^+}f(x)=b<f(c)$ , then all $x>c$ satisfies $f(x)\leq b$ and all $x<c$ satisfies $f(x)\geq f(c)$ so $(b,\,f(c))$ is not in the range of $f$ , contradicting surjectivity. Hence $\lim_{x\to c^+}f(x)=f(c)$ . Clearly we must have $\lim_{x\to\infty}f(x)=0$ , as otherwise a positive number arbitrarily close to $0$ would not be in the range of $f$ . Thus $\lim_{y\to\infty}f(f(xy)+1)=f(1)$ and $\lim_{y\to\infty}xf(x+f(y))=xf(x)$ so $f(x)=\frac{f(1)}{x}$ , as desired. $\square$

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#5 h Apr 26, 2025, 12:47 PM

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Veryyy cool fe

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