Functions

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Potla

1886 posts

Potla

#1 h Feb 21, 2009, 1:05 PM • 2 Y

Y by Adventure10, HWenslawski

Find all functions $f: \mathbb{R}\longrightarrow \mathbb{R}$ such that
$f(x+y)+f(y+z)+f(z+x)\ge 3f(x+2y+3z)$
for all $x, y, z \in \mathbb R$ .

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arcsin1.01

27 posts

arcsin1.01

#2 h Feb 21, 2009, 2:28 PM • 2 Y

Y by Adventure10, HWenslawski

Hint 1
Hint 2
Big hint (do not open before you give up)

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addictedtomath

108 posts

addictedtomath

#3 h Apr 17, 2014, 4:12 PM • 2 Y

Y by Adventure10, HWenslawski

solution

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popcorn1

1098 posts

popcorn1

#4 h Dec 4, 2020, 2:17 AM

Y by

$1 + 2 = 3$ again?!

The answer is $f(x)=c$ for all reals $c$ ; it's easy to see that these solutions work. Let $P(x,y,z)$ denote the given assertion.

Setting $P(x,0,0)$ gives $3f(x) \leq 2f(x) + f(0)$ , so $f(x) \leq f(0)$ for all reals $x$ . Setting $P(\frac{x}{2},\frac{x}{2},-\frac{x}{2})$ gives $3f(0) \leq 2f(0) + f(x)$ , so $f(0) \leq f(x)$ for all reals $x$ . Therefore $f(x) \leq f(0) \leq f(x)$ , or $f$ is constant.

Remark. I really like this FE because both steps are motivated: you get the first from just doing stuff and you wonder, ``huh, how do I use the inequality?'' Then the solution is natural.

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MrOreoJuice

594 posts

MrOreoJuice

#5 h Sep 3, 2021, 10:17 AM • 2 Y

Y by Mango247, Mango247

Really cute

As usual let $P(x,y,z)$ denote the given assertion.

$P(x,0,0) \implies f(x) \le f(0)$ for all $x$ .
$P\left( \dfrac{-3x}{2} , \dfrac{3x}{2} , \dfrac {-x}{2} \right) \implies 2f(0) \le f(-2x) + f(x)$ which is also $\le 2f(0) \implies f(-2x) + f(x) = 2f(0)$ .
$P(-x , 2x , -x) \implies f(x) \ge f(0)$ .

So $f(x) = c$ for constant $c$ .

Edit: wait bruh how did I miss the more natural substitution shown above.

This post has been edited 1 time. Last edited by MrOreoJuice, Sep 3, 2021, 10:21 AM

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jasperE3

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jasperE3

#6 h Sep 3, 2021, 4:42 PM

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$P(x,0,0)\Rightarrow f(x)\le f(0)$
$P(x,x,-x)\Rightarrow f(2x)\ge f(0)\Rightarrow f(x)\ge f(0)\Rightarrow\boxed{f(x)=c}$ for some $c\in\mathbb R$ , which works.

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554183

484 posts

554183

#7 h Oct 14, 2021, 4:35 AM

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$P(y,y,-y) \implies f(2y)+f(0)+f(0) \geq 3f(0) \implies f(x) \geq f(0) \forall x$
$P(-y,y,-y) \implies f(0)+f(0)+f(-2y) \geq 3f(-2y) \implies f(0) \geq f(-2y) \implies f(0) \geq f(x) \forall x$
Therefore, $f(x)=c$ where $c$ is a constant for all $x$

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HamstPan38825

8857 posts

HamstPan38825

#8 h Mar 25, 2023, 2:15 AM

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Only constant functions work.

Summing this inequality cyclically, $f(x+2y+3z) + f(y+2z+3x)+f(z+2x+3y) \leq f(x+y) + f(y+z) + f(z+x).$ On the other hand, substituting $P(2x+y, 2y+z, 2z+x)$ , we obtain exactly the reverse inequality. As a result, equality must hold everywhere. Now, setting $y=z=0$ , $3f(x) = f(x) + 2f(0) \implies f(x) = f(0)$ is constant, as needed.

This post has been edited 1 time. Last edited by HamstPan38825, Mar 25, 2023, 2:15 AM

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Mogmog8

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Mogmog8

#9 h Apr 15, 2023, 4:54 AM • 1 Y

Y by centslordm

We claim the only solution is $f(x)=c$ , where $c$ is a real constant. Note that this satisfies the given conditions. Let $P(x,y,z)$ denote the given assertion and $P(x,0,0)$ yields $f(x)\le f(0)$ . Then, $P(-z,-z,z)$ yields $f(0)\le f(-2z)$ so $f(x)\ge f(0)$ . Hence, $f(x)=f(0)$ so we are done. $\square$

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joshualiu315

2533 posts

joshualiu315

#10 h Oct 17, 2023, 5:54 AM

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Plugging in $(x,y,z)=(x,0,0)$ to get

$3f(x) \le 2f(x)+f(0)$ $\implies f(x) \le f(0)$
Plugging in $(x,y,z)=(x/2,x/2,-x/2)$ gives

$3f(0) \le f(x)+2f(0)$ $\implies f(x) \ge f(0)$
Thus, $f(x)=f(0)$ , so our answer is $\boxed{f(x)=c, \ c \in \mathbb{R}}$ .

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shendrew7

794 posts

shendrew7

#11 h Dec 17, 2023, 1:45 AM

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Substitute $(2k,0,0)$ and $(k,k-k)$ to find
$f(2k) \leq f(0), \quad f(2k) \ge f(0).$
Hence the constant $f(0)$ is the only value $f$ can take, so $\boxed{f(x) = c, \quad c \in \mathbb{R}}$ , which evidently works. $\blacksquare$

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kamatadu

479 posts

kamatadu

#12 h Dec 31, 2023, 6:31 AM

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$f\equiv c$ for some constant $c$ is the solution.

$P(x,0,0) \implies 3f(x) \le f(x) + f(0) + f(x) \implies f(x) \le f(0)$ .

$P(-x,-x,x) \implies 3f(-x + 2(-x) + 3x) \le f(-2x) + f(0) + f(0) \implies 3f(0) \le f(-2x) + 2f(0) \implies f(-2x) \ge f(0)$ .
Now changing $-2x \rightarrow x$ , we get that $f(x) \ge f(0)$ .

Thus $f(x) = f(0)$ . :yoda:

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dolphinday

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dolphinday

#14 h Mar 6, 2024, 8:18 PM

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Letting $y = z = 0$ gives $f(x) \leq f(0)$ .Then $z = -x = -y$ gives $3f(0) \leq f(2x) + 2f(0)$ so $f(0) \leq f(2x)$ , which is only possible if $f(x) = f(0) = c$ .

This post has been edited 1 time. Last edited by dolphinday, Mar 6, 2024, 8:40 PM

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eibc

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eibc

#15 h Mar 11, 2024, 8:26 PM

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The answer is $f \equiv c$ for any real constant $c$ , which works. Let $P(x, y, z)$ denote the given assertion. From $P(x, 0, 0)$ we have $3f(x) \le 2f(x) + f(0)$ , so $f(x) \le f(0)$ . From $P(\tfrac{x}{2}, \tfrac{x}{2}, -\tfrac{x}{2})$ we have $3f(0) \le f(x) + 2f(0)$ , so $f(x) \ge f(0)$ . Hence $f(x) = f(0)$ for all $x$ , which implies the solution set.

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pikapika007

298 posts

pikapika007

#16 h Mar 23, 2024, 5:14 PM

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The answer is $f(x) = c$ only - clearly this works. Now
$P(x, 0, 0) \implies 3f(x) \le 2f(x) + f(0) \implies f(x) \le f(0)$ so
$3f(x+2y+3z) \le f(x+y) + f(y+z) + f(z+x) \le 3f(0).$ Now $P(a, a, -a)$ implies that
$3f(0) \le f(2a) + 2f(0) \le 3f(0)$ so $f(0) \le f(a)$ for all $a$ , and we're done.

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Jndd

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Jndd

#17 h Jul 25, 2024, 7:22 PM

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The answer is $f(x)=c$ for $c\in \mathbb{R}$ , and it is easy to see that this satisfies our inequality.

Plugging in $y=z=0$ , we get $3f(x)\leq 2f(x) + f(0)$ , which gives $f(x)\leq f(0)$ for all $x$ . Then, plugging in $x=y=-z$ , we get $3f(0)\leq f(-2z)+2f(0)$ , so $f(0)\leq f(-2z)$ , giving $f(x)\geq f(0)$ for all $x$ . Since $f(x)\geq f(0)\geq f(x)$ for all $x$ , we must have $f(x)=f(0)$ , as desired.

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Markas

105 posts

Markas

#18 h Aug 20, 2024, 7:55 PM

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Let y = z = 0. Now we plug that in and we get that $3f(x) \leq f(x) + f(0) + f(x)$ $\Rightarrow$ $f(x) \leq f(0)$ . Now let x = -z and y = -z. After we plug that in, we get $3f(0) \leq f(-2z) + f(0) + f(0)$ $\Rightarrow$ $f(0) \leq f(-2z)$ . By the two substitutions we did we get $f(x) \leq f(0) \leq f(-2z)$ and we can choose x = -2z $\Rightarrow$ we now have $f(x) \leq f(0) \leq f(x)$ $\Rightarrow$ $f(0) = f(x)$ $\Rightarrow$ f is constant. Now we only need to check this which is obviously true since $3c \leq c + c + c$ . We are ready since we proved that $f(x) = c$ is the only solution and it works.

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balllightning37

388 posts

balllightning37

#19 h Aug 31, 2024, 8:28 PM

Y by

Neat!

$P(x,0,0)$ implies $3f(x)\leq 2f(x)+f(0)$ or $f(x)\leq f(0)$ for all $x$ .

Then, $P(x,x,-x)$ implies $3f(0)\leq f(x)+2f(0)$ or $f(0)\leq f(x)$ for all $x$ . motivation

Combining, these two, we get that $f(x)=c$ for some constant $c$ , which of course works.

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AshAuktober

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AshAuktober

#20 h Jan 10, 2025, 4:52 PM

Y by

Very kawaii, as expected from a Russian problem!
The only function that works is the constant function. It works as then the inequation becomes $3c \ge 3c$ which is true. We now prove it is the only such function.
Let $P(x, y, z)$ note the given assertion.

Claim 1: $f(x) \le f(0)$ for all $x \in \mathbb{R}$ .
Proof: $P(x, 0, 0)$ $\implies 2f(x) + f(0) \ge 3f(x) \implies f(x) \le f(0)$ as desired. $\square$

Claim 2: $f(x) \ge f(0)$ for all $x \in \mathbb{R}$ .
Proof: $P\left(\frac x2, \frac x2, -\frac x2 \right)$ $\implies 2f(0) + f(x) \ge 3f(0) \implies f(x) \ge f(0),$ as desired. $\square$

Combining the above two claims, $f(x) = f(0)$ , and thus $f$ is indeed a constant function. $\blacksquare$

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eg4334

637 posts

eg4334

#21 h Feb 17, 2025, 3:35 AM

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Let $x+y=a, y+z=b, x+z=c$ so it rewrites into $f(a)+f(b)+f(c) \geq 3f(2b+c)$ . Set $b=0, a=c$ to get $2f(a)+f(0) \geq 3f(a) \implies f(0) \geq f(a)$ And then $b=c=0$ to get $f(a) \geq f(0)$ . Therefore $f(a)=f(0)$ so $\boxed{f(x) \equiv c}$

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blueprimes

344 posts

blueprimes

#22 h Feb 26, 2025, 2:02 AM

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We claim the answer is $f(x) = c$ for a constant $c$ . Obviously it works. Now the assertion $y = z = 0$ yields $f(x) \ge f(0)$ for all $x$ , whereas $(x, y, z) \mapsto \left(\dfrac{x}{2}, \dfrac{x}{2}, - \dfrac{x}{2} \right)$ yields $f(x) \le f(0)$ for all $x$ . So $f(x) = f(0)$ and $f$ is constant as wanted.

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Marcus_Zhang

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Marcus_Zhang

#23 h Mar 13, 2025, 11:19 PM

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Almost a one liner

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Maximilian113

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Maximilian113

#24 h Mar 16, 2025, 4:00 PM

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Let $P(x, y, z)$ denote the assertion. Then $P(-x, -x, x) \implies 3f(0) \leq f(-2x)+2f(0) \implies f(x) \geq f(0) \, \forall x \in \mathbb R.$ Also, $P(-x, x, -x) \implies 3f(-2x) \leq 2f(0)+f(-2x) \implies f(x) \leq f(0) \, \forall x \in \mathbb R.$ It follows that $f(x)=c$ for some constant $c,$ and this clearly satisfies the assertion.

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Ilikeminecraft

609 posts

Ilikeminecraft

#25 h Apr 25, 2025, 8:06 AM

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I claim that $f(x) = c$ for some constant $c\in\mathbb R$ . This clearly works.

Plug in $x = y = -z$ to get that $f(0) \leq f(2x),$ and then $x = -y = z$ to get that $f(2x) \leq f(0).$ Thus, $f(x)$ is constant.

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