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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
0 replies
Need help with barycentric
Sadigly   0
12 minutes ago
Hi,is there a good handout/book that explains barycentric,other than EGMO?
0 replies
Sadigly
12 minutes ago
0 replies
Combinatorics
P162008   3
N 36 minutes ago by P162008
Let $m,n \in \mathbb{N}.$ Let $[n]$ denote the set of natural numbers less than or equal to $n.$

Let $f(m,n) = \sum_{(x_1,x_2,x_3, \cdots, x_m) \in [n]^{m}} \frac{x_1}{x_1 + x_2 + x_3 + \cdots + x_m} \binom{n}{x_1} \binom{n}{x_2} \binom{n}{x_3} \cdots \binom{n}{x_m} 2^{\left(\sum_{i=1}^{m} x_i\right)}$

Compute the sum of the digits of $f(4,4).$
3 replies
P162008
4 hours ago
P162008
36 minutes ago
Find min and max
lgx57   0
an hour ago
Source: Own
$x_1,x_2, \cdots ,x_n\ge 0$,$\displaystyle\sum_{i=1}^n x_i=m$. $k_1,k_2,\cdots,k_n >0$. Find min and max of
$$\sum_{i=1}^n(k_i\prod_{j=1}^i x_j)$$
0 replies
lgx57
an hour ago
0 replies
Find min
lgx57   0
an hour ago
Source: Own
$a,b>0$, $a^4+ab+b^4=60$. Find min of
$$4a^2-ab+4b^2$$
$a,b>0$, $a^4-ab+b^4=60$. Find min of
$$4a^2-ab+4b^2$$
0 replies
lgx57
an hour ago
0 replies
No more topics!
A three-variable functional inequality on non-negative reals
Tintarn   11
N Apr 12, 2025 by jasperE3
Source: Dutch TST 2024, 1.2
Find all functions $f:\mathbb{R}_{\ge 0} \to \mathbb{R}$ with
\[2x^3zf(z)+yf(y) \ge 3yz^2f(x)\]for all $x,y,z \in \mathbb{R}_{\ge 0}$.
11 replies
Tintarn
Jun 28, 2024
jasperE3
Apr 12, 2025
A three-variable functional inequality on non-negative reals
G H J
Source: Dutch TST 2024, 1.2
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Tintarn
9042 posts
#1 • 3 Y
Y by ehuseyinyigit, Sedro, Parsia--
Find all functions $f:\mathbb{R}_{\ge 0} \to \mathbb{R}$ with
\[2x^3zf(z)+yf(y) \ge 3yz^2f(x)\]for all $x,y,z \in \mathbb{R}_{\ge 0}$.
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Hamzaachak
61 posts
#2 • 1 Y
Y by JanHaj
The anwser is f(x) = cx^2 (with c is a positif integer or 0)
This post has been edited 1 time. Last edited by Hamzaachak, Jun 29, 2024, 3:25 PM
Reason: Mousse trap
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ACGNmath
327 posts
#3
Y by
We claim that the answer is $\boxed{f(x) = cx^2}$ with $c \geq 0$ a nonnegative real. To see that this works, apply AM-GM:
\[2x^3 z^3 + y^3 = x^3 z^3 + x^3 z^3 + y^3 \geq 3x^2 yz^2.\]We now show that this is the only family of solutions. Let $P(x,y,z)$ denote the proposition $2x^2z f(z) + yf(y) \geq 3yz^2 f(x)$.

$P(0,1,z): f(1) \geq 3z^2 f(0)$. Since $z$ is arbitrary, this means that $f(0)=0$.

$P(x,x,1): 2x^3 f(1) + xf(x) \geq 3xf(x) \quad\Rightarrow\quad 2x^3 f(1) \geq 2xf(x)\quad\Rightarrow\quad f(x) \leq cx^2$
where $c = f(1)$.

$P(1,x,x): 2x f(x) + xf(x) \geq 3cx^3 \quad\Rightarrow\quad f(x) \geq cx^2$.

Combining the two inequalities, we get $f(x) = cx^2$ as desired.
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ayeen_izady
32 posts
#4
Y by
Claim: $f(x)=cx^2$ for some $c\ge 0$.
Proof: Let $f(x)=x^2g(x)$. Note that $2x^3z^3g(z)+y^3g(y)\ge 3yx^2z^2g(x)$. By putting $y=xy$ we get that $x^3z^3g(xz)\ge x^3z^3g(x)$. If $x,z>0$ it's clear that $g(x)=c$ for all positive reals $x$ because $g(t)\ge g(s)$ and $g(s)\ge g(t)$ for all positive reals $s,t$. Thus $f(x)=cx^2$ for all positive reals $x$. You can now get that $f(0)=0$ and hence $f(x)=cx^2$
This post has been edited 1 time. Last edited by ayeen_izady, Jun 29, 2024, 2:10 PM
Reason: typo
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Sedro
5845 posts
#5
Y by
ACGNmath wrote:
$P(0,1,z): f(1) \geq 3z^2 f(0)$. Since $z$ is arbitrary, this means that $f(0)=0$.
How do we conclude this? Can't $f(0)$ be negative?
This post has been edited 1 time. Last edited by Sedro, Jun 29, 2024, 2:56 PM
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Hamzaachak
61 posts
#6
Y by
Sedro wrote:
ACGNmath wrote:
$P(0,1,z): f(1) \geq 3z^2 f(0)$. Since $z$ is arbitrary, this means that $f(0)=0$.
How do we conclude this? Can't $f(0)$ be negative?

y=z=0 give f(0) non negatif
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jenishmalla
5 posts
#7
Y by
This was pretty fun and fell into place pretty quick
Motivations for substitution:mostly getting things to add and subtract and making expression less ugly
\[
\text{Let } P(x, y, z) \text{ be the given assertion.}
\]
\[
P(y, y, 1) \Rightarrow y^2 f(1) \geq f(y) \tag{1}
\]\[
P(1, y, y) \Rightarrow f(y) \geq y^2 f(1) \tag{2}
\]
\[
\text{From (1) and (2), we get:} \quad f(y) = y^2 f(1)
\]
\[
\Rightarrow f(x) = kx^2 \quad \text{where } k = f(1)
\]
\[
\text{Now, plug into the original inequality:}
\]\[
k(x^3 z^3 + x^3 z^3 + y^3) \geq k (3x^3 y z^3)
\]
\[
\Rightarrow \text{Inequality holds for all } x, y, z  \text{ if } k \geq 0
\]
\[
\boxed{f(x) = kx^2 \text{ for all } x \text{ where } k \geq 0 \text{ with equality when } x = y = z \text{ or } k = 0}
\]
(sorry for the bad latex)
This post has been edited 2 times. Last edited by jenishmalla, Apr 9, 2025, 4:25 PM
Reason: typo and equality case
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ErTeeEs06
64 posts
#8
Y by
hmm so far there is no completely correct solution in this thread
Sedro is right, f(0) can indeed be any value <=0
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jasperE3
11301 posts
#9
Y by
Isn't it $f(x)=cx^2$ for all $x>0$ and $f(0)\le0$?
adapt something like the solution in #3
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ErTeeEs06
64 posts
#10
Y by
jasperE3 wrote:
Isn't it $f(x)=cx^2$ for all $x>0$ and $f(0)\le0$?
adapt something like the solution in #3

Yeah this is indeed correct.
$f(x)=cx^2$ for all $x>0$ and $f(0)=d$ where $c, d$ are constants with $c\geq 0$ en $d\leq 0$
During TST I was the only one who scored 7 on this cause lot of people didn't see the codomain was $\mathbb{R}$ and not $\mathbb R_{\geq 0}$ and lost a point for that
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Alidq
32 posts
#11 • 1 Y
Y by ErTeeEs06
If we plug $x=0$ and $y=1$ we have $f(1) \geq 3f(0)z^2$, $\forall z\geq 0$. Now if $f(0)>0$ the the RHS is unbounded so we have a contradiction. Thus $f(0) \leq 0$. Substituing $z=0$ we have $yf(y) \geq 0$ so $f(y) \geq 0$ for all $y > 0$ if we take $y=1$ we have $f(1) \geq 0$. Substituting $x=y$ and $z=1$ yields $$2y^3f(1)+yf(y)\geq 3yf(y)$$, so we have $f(y) \leq y^2f(1)$ ,$y>0$ $(1)$. Now if we take $z=y$ and $x=1$ we obtain $$2yf(y)+yf(y) \geq 3y^3f(1)$$so $f(y)\geq y^2f(1)$ , $y>0$ $(2)$.From $(1)$ and $(2)$ we have $f(y)=y^2f(1)$ , $y>0$. Thus $f(x)=kx^2$ where $k>0$ and $f(0)=l$ where $l \leq 0$
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jasperE3
11301 posts
#12
Y by
ErTeeEs06 wrote:
jasperE3 wrote:
Isn't it $f(x)=cx^2$ for all $x>0$ and $f(0)\le0$?
adapt something like the solution in #3

Yeah this is indeed correct.
$f(x)=cx^2$ for all $x>0$ and $f(0)=d$ where $c, d$ are constants with $c\geq 0$ en $d\leq 0$
During TST I was the only one who scored 7 on this cause lot of people didn't see the codomain was $\mathbb{R}$ and not $\mathbb R_{\geq 0}$ and lost a point for that

good job
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