A three-variable functional inequality on non-negative reals

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A three-variable functional inequality on non-negative reals

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Source: Dutch TST 2024, 1.2

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#1 h Jun 28, 2024, 3:31 PM • 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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#2 h Jun 29, 2024, 1:06 PM • 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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#3 h Jun 29, 2024, 2:04 PM

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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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#4 h Jun 29, 2024, 2:09 PM

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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

#5 h Jun 29, 2024, 2:56 PM

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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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#6 h Jun 29, 2024, 3:29 PM

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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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#7 h Apr 9, 2025, 4:14 PM

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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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#8 h Apr 9, 2025, 6:03 PM

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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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#9 h Apr 9, 2025, 7:06 PM

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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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#10 h Apr 10, 2025, 7:03 AM

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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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#11 h Apr 11, 2025, 9:52 PM • 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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#12 h Apr 12, 2025, 1:18 AM

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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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