real+ FE

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

real+ FE

G H J

Reveal topic tags

functional equation

L

G H BBookmark kLocked kLocked NReply

Source: Own, PDC001-P7

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

24 posts

#1 h Apr 21, 2025, 11:24 AM

Y by

Let $f : \mathbb{R}^+ \to \mathbb{R}^+$ be a function such that
$f(x)f(x^2 + y f(y)) = f(x)f(y^2) + x^3$ for all $x, y \in \mathbb{R}^+$ . Determine all such functions $f$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

79 posts

#2 h Apr 21, 2025, 2:11 PM

Y by

We'll prove $f(x)=x$ is the only solution. Suppose $\exists u: f(u)<u$ . Setting $x= \sqrt{u^2-uf(u)},y=u$ we get $f(x)f(u^2)=f(x)f(x^2+yf(y))=f(x)f(u^2)+x^3 \Rightarrow x^3=0$ Which is a contradiction. and so $f(x)\ge x$ for all $x$ . Then using this we get $f(x)x^2+yf(x)f(y)\le f(x)f(y^2)+x^3 \Rightarrow x^2(1-\frac{x}{f(x)}) \le f(y^2)-yf(y) \Rightarrow \forall y: f(y^2)\ge yf(y)$ Let $g(x)=\frac{f(x)}{x}$ . Let $x>1$ , $x_i = x^{2^i}$ and $c>1$ be a constant. If there exists infinitely many indices $i$ such that $g(x_i)\ge c$ , for an arbitrary $y$ , we get $g(x_i) \ge c \Rightarrow (x_i)^2(1-\frac{1}{c})\le (x_i)^2(1-\frac{1}{g(x_i)}) \le f(y^2)-yf(y)$ But the left hand side is unbounded which is contradiction. And so for every $c>1$ , there exists an integer $N$ s.t. $\forall n>N: g(x_n) < c$ which gives $\lim g(x_n) =1$ but we already had $g(x_n)\ge g(x)$ for all $x$ . this gives $\forall x>1: g(x) \le \lim g(x_n) = 1$ . we also had $g(x)\ge 1$ and so $g(x)=1$ for all $x>1$ . setting $y$ great enough gives $f(x)=x$ which works.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

10 posts

#3 h Apr 21, 2025, 2:29 PM

Y by

Denote the given assertion by $P(x,y).$

From $P(x,y)$ we have $f(x)(f(x^2+yf(y))-f(y^2))=x^3$ . If $f(y)<y$ for some $y$ then by plugging $x=\sqrt{y^2-yf(y)}$ we have $0=x^3$ . A contradiction. Which means $f(y)\ge y$ for all $y$ .

$P(1,1) : f(1)^2+1=f(1)f(f(1)+1)\ge f(1)(f(1)+1) \Longrightarrow 1\ge f(1)$ . Thus, $f(1)=1$ .

$P(x,1) : f(x)+x^3=f(x)f(x^2+1)\ge (x^2+1)f(x) \Longrightarrow x^3\ge x^2f(x)\Longrightarrow x\ge f(x)$ . Thus, $f(x)=x$ $\forall x \in \mathbb{R^+}$ which clearly fits.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1556 posts

#4 h Apr 22, 2025, 5:08 AM

Y by

Denote $P(x,y)$ the assertion of the given F.E.
Claim 1: $f(x) \ge x$ for all positive reals $x$ .
Proof: Suppose FTSOC that $v>f(v)$ for some $v$ then from $P (\sqrt{v^2-vf(v)},v )$ we get can get that $(v^2-vf(v))^{\frac{3}{2}}=0$ which is clearly a contradiction.
Claim 2: $f(x^2) \ge xf(x)$ for all positive reals $x$
Proof: Notice from $P(x,y)$ and Claim 1 that:
$x^3+f(x)f(y^2) \ge x^2f(x)+yf(x)f(y) \ge x^3+yf(x)f(y) \implies f(y^2) \ge yf(y)$ Claim 3: $f(1)=1$ .
Proof: We have $f(1) \ge 1$ from Claim 1 but also $P(1,1)$ gives $f(1)f(1+f(1))=f(1)^2+1$ and therefore $f(1)^2+1 \ge f(1)^2+f(1)$ so $1 \ge f(1)$ so $f(1)=1$ as desired.
The finish: From $P(x,1)$ and ineqs notice that $x^3+f(x) \ge x^2f(x)+f(x)$ so $x \ge f(x)$ so $f(x)=x$ for all positive reals $x$ thus we are done :cool:

:cool:

.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

11385 posts

#5 h Apr 22, 2025, 6:09 PM

Y by

pomodor_ap wrote:

Let $f : \mathbb{R}^+ \to \mathbb{R}^+$ be a function such that
$f(x)f(x^2 + y f(y)) = f(x)f(y^2) + x^3$ for all $x, y \in \mathbb{R}^+$ . Determine all such functions $f$ .

Very nice.

Claim 1: $f(x)\ge x$
Suppose $f(y)<y$ for some $y$ . Setting $x=\sqrt{y^2-yf(y)}$ so that $f(x)f(x^2+yf(y))=f(x)f\left(y^2\right)$ , we get:
$\left(\sqrt{y^2-yf(y)}\right)^3=0\Rightarrow f(y)=y,$ contradiction.

Now taking $x\mapsto\sqrt x$ , $y\mapsto\sqrt y$ in the original equation, we get:
$f\left(x+\sqrt yf\left(\sqrt y\right)\right)=\frac{x^{3/2}}{f\left(\sqrt x\right)}+f(y),$ so defining $g(x)=\sqrt xf\left(\sqrt x\right)$ and $h(x)=\frac{x^{3/2}}{f\left(\sqrt x\right)}$ , we have:
$f(x+g(y))=h(x)+f(y).$ Note that $f(1)=g(1)$ , and by Claim 1, $h(x)\le x$ .

Claim 2: $g(x)=f(x)$
First, we have:
$f(g(x)+g(y))=h(g(x))+f(y)$ and switching $x,y$ , we get that $h(g(x))=f(x)+c$ for some constant $c\in\mathbb R$ .
Since $h(x)\le x$ this gives $g(x)\ge f(x)+c$ .

So this equation becomes $f(g(x)+g(y))=f(x)+f(y)+c$ , and since $f(x)\ge x$ we have:
$g(x)+g(y)\le f(x)+f(y)+c.$ Setting $y=1$ we get $g(x)\le f(x)+c$ .

Because of these two, $g(x)=f(x)+c$ for all $x\in\mathbb R^+$ . Since $g(1)=f(1)$ again, we have $c=0$ , hence proven.

Claim 3: $h(x)=x$
Trivial:
$x+f(y)\le f(x+f(y))=f(x+g(y))=h(x)+f(y)\le x+f(y)$ with equality only when $h(x)=x$ .

Then $\frac{x^{3/2}}{f\left(\sqrt x\right)}=x$ , giving the unique solution $\boxed{f(x)=x}$ which fits.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a