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Benelux fe
ErTeeEs06   0
4 minutes ago
Source: BxMO 2025 P1
Does there exist a function $f:\mathbb{R}\to \mathbb{R}$ such that $$f(x^2+f(y))=f(x)^2-y$$for all $x, y\in \mathbb{R}$?
0 replies
ErTeeEs06
4 minutes ago
0 replies
$5^t + 3^x4^y = z^2$
Namisgood   0
14 minutes ago
Source: JBMO shortlist 2017
Solve in nonnegative integers the equation $5^t + 3^x4^y = z^2$
0 replies
Namisgood
14 minutes ago
0 replies
Concurrent lines
MathChallenger101   2
N 19 minutes ago by pigeon123
Let $A B C D$ be an inscribed quadrilateral. Circles of diameters $A B$ and $C D$ intersect at points $X_1$ and $Y_1$, and circles of diameters $B C$ and $A D$ intersect at points $X_2$ and $Y_2$. The circles of diameters $A C$ and $B D$ intersect in two points $X_3$ and $Y_3$. Prove that the lines $X_1 Y_1, X_2 Y_2$ and $X_3 Y_3$ are concurrent.
2 replies
MathChallenger101
Feb 8, 2025
pigeon123
19 minutes ago
Find all natural numbers $n$
ItsBesi   7
N 24 minutes ago by justaguy_69
Source: Kosovo Math Olympiad 2025, Grade 9, Problem 2
Find all natural numbers $n$ such that $\frac{\sqrt{n}}{2}+\frac{10}{\sqrt{n}}$ is a natural number.
7 replies
ItsBesi
Nov 17, 2024
justaguy_69
24 minutes ago
diopantine 5^t + 3^x4^y = z^2 , solve in nonnegative
parmenides51   10
N 41 minutes ago by Namisgood
Source: JBMO Shortlist 2017 NT4
Solve in nonnegative integers the equation $5^t + 3^x4^y = z^2$.
10 replies
parmenides51
Jul 25, 2018
Namisgood
41 minutes ago
\pi(n) is a good divisor
sefatod628   1
N 42 minutes ago by kiyoras_2001
Source: unknown/given at a math club
Prove that there is an infinity of $n\in \mathbb{N^*}$ such that $\pi(n) \mid n$.

Hint :Click to reveal hidden text
1 reply
sefatod628
Yesterday at 2:37 PM
kiyoras_2001
42 minutes ago
Easy Functional Inequality Problem in Taiwan TST
chengbilly   3
N an hour ago by shanelin-sigma
Source: 2025 Taiwan TST Round 3 Mock P4
Let $a$ be a positive real number. Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that $af(x) - f(y) + y > 0$ and
\[
    f(af(x) - f(y) + y) \leq x + f(y) - y, \quad \forall x, y \in \mathbb{R}^+.
    \]
proposed by chengbilly
3 replies
chengbilly
4 hours ago
shanelin-sigma
an hour ago
4 var inequality
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b,c,d >0 $ and $ abcd=3,a+b+c+d=6 $. Prove that
$$ a^2+b^2+c^2+d^2  \leq 12 $$$$   a^3+b^3+c^3+d^3 \leq 30$$$$   a^4+b^4+c^4+d^4 \leq84$$
1 reply
sqing
an hour ago
sqing
an hour ago
Dou Fang Geometry in Taiwan TST
Li4   1
N 2 hours ago by Parsia--
Source: 2025 Taiwan TST Round 3 Mock P2
Let $\omega$ and $\Omega$ be the incircle and circumcircle of the acute triangle $ABC$, respectively. Draw a square $WXYZ$ so that all of its sides are tangent to $\omega$, and $X$, $Y$ are both on $BC$. Extend $AW$ and $AZ$, intersecting $\Omega$ at $P$ and $Q$, respectively. Prove that $PX$ and $QY$ intersects on $\Omega$.

Proposed by kyou46, Li4, Revolilol.
1 reply
Li4
Today at 5:03 AM
Parsia--
2 hours ago
4 var inequality
sqing   1
N 2 hours ago by sqing
Source: https://bbs.emath.ac.cn/thread-39778-1-1.html
Let $ a,b,c,d>0 $ and $ a+b+c+d=4. $ Prove that$$a\sqrt{bc}+b\sqrt{cd}+c\sqrt{da}+d\sqrt{ab}\leq 2(1+\sqrt{abcd})$$Let $ a,b,c,d\geq -1 $ and $ a+b+c+d=0. $ Prove that$$ab+bc+cd\leq \frac{5}{4}$$
1 reply
sqing
2 hours ago
sqing
2 hours ago
easy functional
B1t   5
N 2 hours ago by Haris1
Source: Mongolian TST 2025 P1.
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[
f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]
5 replies
B1t
4 hours ago
Haris1
2 hours ago
Two Functional Inequalities
Mathdreams   7
N Apr 7, 2025 by John_Mgr
Source: 2025 Nepal Mock TST Day 2 Problem 2
Determine all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(x) \le x^3$$and $$f(x + y) \le f(x) + f(y) + 3xy(x + y)$$for any real numbers $x$ and $y$.

(Miroslav Marinov, Bulgaria)
7 replies
Mathdreams
Apr 6, 2025
John_Mgr
Apr 7, 2025
Two Functional Inequalities
G H J
Source: 2025 Nepal Mock TST Day 2 Problem 2
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Mathdreams
1469 posts
#1 • 4 Y
Y by PikaPika999, AlexCenteno2007, cubres, khan.academy
Determine all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(x) \le x^3$$and $$f(x + y) \le f(x) + f(y) + 3xy(x + y)$$for any real numbers $x$ and $y$.

(Miroslav Marinov, Bulgaria)
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grupyorum
1413 posts
#2 • 2 Y
Y by PikaPika999, cubres
I will show $f(x)=x^3$ for all $x$, which clearly works.

Taking $x=y=0$ in both, we find $f(0)\le 0$ and $f(0)\ge 0$, so $f(0)=0$. Taking $y=-x$ in the second, we find $0=f(0)\le f(x)+f(-x)$. Likewise, $f(x)\le x^3$ and $f(-x)\le (-x)^3=-x^3$. So, $f(x)+f(-x)\le 0$. Combining, $f(x)+f(-x)=0$ for all $x$. Lastly, using $f(-x)=-f(x)$, we get $-x^3 \ge f(-x)=-f(x)$, so $f(x)\ge x^3$ too. Thus, $f(x)=x^3$ for all $x$.
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kokcio
69 posts
#3 • 2 Y
Y by PikaPika999, cubres
Putting $x=y=0$, we have $f(0)\leq 2f(0)$, but $f(0)\leq 0$, so we have to have that $f(0)=0$.
Putting now $y=-x$, we have $0\leq f(x)+f(-x) \leq x^3 + (-x)^3 = 0$, but this means that we have to have $f(x)=x^3$, because of inequalities.
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John_Mgr
67 posts
#4 • 2 Y
Y by PikaPika999, cubres
Claim: $f(x) \equiv x^3$ is the only function $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfies the given conditions.
Let P(x,y) be the assertion for the convenience.
i.e $f(x+y) \le f(x)+f(y)+3xy(x+y)$
P(0,0): $f(0)\ge0$ and $f(0)\le0$, $\Rightarrow$ $f(0)=0$,
Assume $f(x)=g(x)+x^3$, $\implies$ $g(x+y)\le g(x)+g(y)$
This tells us that g is subadditive.
$g(x)\le 0$ as $f(x)\le x^3$
Let Q(x,y) be the assertion for $g(x+y)\le g(x)+g(y)$
Q(x,-x): $g(0)\le g(x)+g(-x)$$\rightarrow g(x)+g(-x)\ge 0$ as $f(0)=0$ but $g(x)+g(-x)\le 0$
So, $g(x)+g(-x)=0$$\Rightarrow g(-x)=-g(x) \ge 0$
The only way both $g(x)\le x$ and $g(-x)\ge 0$ hold is if $g(x)=0$, $\forall$ $x\in \mathbb{R}$
Therefore $\boxed{f(x)=x^3}$, $\forall$ $x\in \mathbb{R}$
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jasperE3
11251 posts
#5 • 5 Y
Y by PikaPika999, Assassino9931, megarnie, navier3072, cubres
Let $g(x)=f(x)-x^3+x$, it reduces to this old problem:
https://artofproblemsolving.com/community/c6h257315p1403063
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Maximilian113
561 posts
#6 • 2 Y
Y by PikaPika999, cubres
Note that $x=y=0 \implies 0 \leq f(0) \leq 0 \implies f(0)=0.$ Now $$y=-x \implies 0 \leq f(x)+f(-x) \leq f(x) -x^3 \implies x^3 \leq f(x) \implies f(x)=x^3$$for all $x,$ which clearly works.
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Assassino9931
1252 posts
#7 • 1 Y
Y by cubres
jasperE3 wrote:

Nice catch haha
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John_Mgr
67 posts
#8 • 1 Y
Y by cubres
jasperE3 wrote:

Good one!!
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