I did not know that Anton makes FEs as well

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

I did not know that Anton makes FEs as well

G H J

Reveal topic tags

functional equation

Functional inequality

L

G H BBookmark kLocked kLocked NReply

Source: Ukrainian Mathematical Olympiad 2025. Day 1, Problem 11.3

The post below has been deleted. Click to close.

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

77 posts

#1 h Mar 13, 2025, 11:37 PM • 1 Y

Y by Supercali

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for any real numbers $x$ and $y$ , the following inequality holds:
$f\left(x^2+2y f(x)\right) + (f(y))^2 \leq f\left((x+y)^2\right)$ Proposed by Anton Trygub

This post has been edited 4 times. Last edited by mshtand1, Mar 14, 2025, 1:23 AM

Z K Y

The post below has been deleted. Click to close.

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

5538 posts

#2 h Mar 14, 2025, 4:13 AM

Y by

Solved with hgomamogh.

The only solutions are $f\equiv 0$ and $f(x) = x$ , which obviously work. Now we show they are the only solutions.

Let $P(x,y)$ be the given assertion.

$P(0,0): f(0)^2 \le 0 \implies f(0) = 0$ .

$P(0,y): f(y)^2 \le f(y^2)$ . This implies $f(x) \ge 0 \forall x \ge 0$ .

Claim: If $f(x) =0$ , then $f(x^2) = 0$ and $f(-x) = 0$ .
Proof: Note that if $f(x) = 0$ , then $f(x^2) + f(y)^2 \le f((x+y)^2)$ . Setting $y = -x$ gives $f(x^2) + f(-x)^2 \le 0\implies f(x^2) \le 0$ . But since $f(x^2) \ge 0$ also, $f(x^2) = 0$ . To see that $f(-x) = 0$ , note that $f(-x)^2 \le f(x^2) = 0$ , so $f(-x) = 0$ must hold. $\square$

Claim: If there exists a constant $c>0$ with $f(c) = 0$ , then $f(x) = 0$ for all nonnegative $x \le c$ .
Proof: For any nonnegative $x \le c$ , $P(c-x, x)$ gives that $f((c-x)^2 + 2x f(c-x)) + f(x)^2 \le 0$ . However, since $(c-x)^2, x, c - x$ are nonnegative, $f((c-x)^2 + 2xf(c-x)) \ge 0$ , so $f((c-x)^2 + 2x f(c-x)) = f(x) = 0$ must hold. $\square$

Claim: If $f(a) = 0$ , then $f(2a) = 0$ .
Proof: Fix $a$ with $f(a) = 0$ . Note that $f(a^2) = f(-a) = 0$ .

$P(-a,2a): f(a^2 + 4a f(-a)) + f(2a)^2 \le 0$ . However, the LHS becomes $f(2a)^2$ . So $f(2a)^2 \le 0 \implies f(2a) = 0$ . $\square$

This means that if we have some nonzero root of $f$ , by the first claim, we can find some positive number $c$ with $f(c) = 0$ , and then we note that every nonnegative real less than or equal to $c$ is a root of $f$ . But then we can get that $2^n c$ for any integer $n$ is a root of $f$ , proving that all nonnegative reals are roots of $f$ , so $f\equiv 0$ . Now assume otherwise (so $f$ is injective at $0$ ).

Claim: $f(2f(x) - 2x) = 0$ .
Proof: If we set $y = 2f(x) - 2x$ , we get $f(2f(x) - 2x)^2 \le 0$ , so $f(2f(x) - 2x) = 0$ . $\square$

By injectivity, we have $2f(x) = 2x \implies f(x) = x$ , so $f$ is the identity, as desired.

Z K Y

The post below has been deleted. Click to close.

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

1460 posts

#3 h Mar 14, 2025, 4:19 AM

Y by

Denote $P(x,y)$ the assertion of the given F.E.
$P(x,0)$ gives $f(0)^2 \le 0$ which gives $f(0)=0$ , now $P(0,x)$ gives $f(x)^2 \le f(x^2)$ which shows that $f$ on non-negative reals takes non-negative real values.
Now from $P(x, 2f(x)-2x)$ we get that $f(2f(x)-2x)=0$ for all reals $x$ and thus if there existed some $c \ne 0$ such that $f(c)=0$ then $f(-2c)=0$ , but also notice we have from $P(c,-c)$ that $f(c^2)=0=f(-c)$ and therefore the sign on the zeroes does not in fact matter so wlog $c>0$ .
Now from $P(c-x,x)$ for $c \ge x \ge 0$ gives that $f(x)=0$ but as we have seen before we can in fact have that if $f(c)=0$ then $f(2^n c)=0$ which means $f(x)=0$ for all $x \ge 0$ and this means that $0 \ge f(x)^2$ so $f(x)=0$ for all reals $x$ or $f$ is injective at $0$ .
On the later case we get $f(x)=x$ for all reals $x$ , so in all cases 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.

ohiorizzler1434

723 posts

ohiorizzler1434

#4 h Mar 14, 2025, 4:25 AM

Y by

What a devious antiproblem written by master troll Anton!

Z K Y

The post below has been deleted. Click to close.

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

148 posts

#5 h Mar 20, 2025, 6:08 AM

Y by

U

mshtand1 wrote:

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for any real numbers $x$ and $y$ , the following inequality holds:
$f\left(x^2+2y f(x)\right) + (f(y))^2 \leq f\left((x+y)^2\right)$ Proposed by Anton Trygub

Nice FE
Ukraine’s FE always have high qualities :love:

:love:

Attachments:

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a