f(x)+f(x+y) \leq f(xy)+f(y)

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f(x)+f(x+y) \leq f(xy)+f(y)

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Source: Estonia TST 2022

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#1 h Jul 12, 2023, 6:29 AM

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Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy the following condition for any real numbers $x{}$ and $y$ $f(x)+f(x+y) \leq f(xy)+f(y).$

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pco

#2 h Jul 12, 2023, 7:16 AM • 1 Y

Y by PRMOisTheHardestExam

augustin_p wrote:

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy the following condition for any real numbers $x{}$ and $y$ $f(x)+f(x+y) \leq f(xy)+f(y).$

Let $P(x,y)$ be the assertion $f(x)+f(x+y)\le f(xy)+f(y)$
Let $a=f(0)$

$P(x,0)$ $\implies$ $f(x)-a\le 0$ $\forall x$
$P(x,-x)$ $\implies$ $f(x)-f(-x)\le f(-x^2)-a\le 0$
$P(-x,x)$ $\implies$ $f(-x)-f(x)\le f(-x^2)-a\le 0$
And so $f(-x^2)=a$ and $f(x)=f(-x)$ $\forall x$

And so $\boxed{f(x)=a\quad\forall x}$ , which indeed fits whatever is $a\in\mathbb R$

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F10tothepowerof34

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F10tothepowerof34

#3 h Aug 8, 2023, 7:48 PM

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As usual let $P(x,y):=f(x)+f(x+y)\le f(xy)+f(y)$

$P(x,0)$ yields $2f(x)\le2f(0)\Longrightarrow f(x)\le f(0)$
$P(-x^2,0)$ yields $2f(-x^2)\le2f(0)\Longrightarrow f(-x^2)\le f(0)$ (!!)
$P\left(\frac{x^4}{4},0\right)$ yields $f\left(\frac{x^4}{4}\right)\le f(0)$ (!!)
Furthermore $P(x,-x)$ yields $f(x)+f(0)\le f(-x^2)+f(-x)$ and $P(-x,x)$ yields $f(-x)+f(0)\le f(-x^2)+f(x)$
Now, notice that from $P(x,-x)+P(-x,x)$ we obtain $f(x)+f(0)+f(-x)+f(0)\le f(-x^2)+f(-x)+f(-x^2)+f(x)\Longrightarrow f(x)+f(-x)+2f(0)\le 2f(-x^2)+f(-x)+f(x)\Longleftrightarrow f(0)\le f(-x^2)$ However, notice that from (!!) we have that $f(0)\le f(-x^2)\le f(0)$ , which forces $f(-x^2)=f(0)\Longrightarrow f(\text{negative})=f(0)$
Furthermore notice that from $P\left(-\frac{x^2}{2},-\frac{x^2}{2}\right)$ we obtain $f(-x^2)=f(0)\le f\left(\frac{x^4}{4}\right)$ , however from (!!) we have that $f(0)\le f\left(\frac{x^4}{4}\right)\le f(0)$ which forces $f\left(\frac{x^4}{4}\right)=f(0)\Longleftrightarrow f(\text{positive})=f(0)$ thus $f(x)=f(0), \forall x\in\mathbb{R}$

In conclusion $\boxed{f(x)=c, \forall x\in\mathbb{R}\text{ and any constant }c\in\mathbb{R}}$ $\blacksquare$ .

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#4 h Aug 28, 2023, 9:39 PM

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augustin_p wrote:

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy the following condition for any real numbers $x{}$ and $y$ $f(x)+f(x+y) \leq f(xy)+f(y).$

$\color{blue}\boxed{\textbf{Answer: }f\equiv c}$
$\color{blue}\boxed{\textbf{Proof:}}$
$\color{blue}\rule{24cm}{0.3pt}$
$f(x)+f(x+y) \leq f(xy)+f(y)...(\alpha)$ In $(\alpha) y\to 0:$
$\Rightarrow 2f(x)\leq 2f(0)$ $\Rightarrow f(x)\leq f(0)...(\beta)$ In $(\alpha) y\to -x:$
$\Rightarrow f(x)+f(0)\leq f(-x^2)+f(-x)...(i)$ By $(\beta):$
$\Rightarrow f(x)+f(-x^2)\leq f(x)+f(0) \leq f(-x^2)+f(-x)$ $\Rightarrow f(x)\leq f(-x)...(\theta)$ In $(\theta) x\to -x:$
$\Rightarrow f(-x)\leq f(x)$ $\Rightarrow f(x)=f(-x)...(\omega)$ Replacing $(\omega)$ in $(i):$
$\Rightarrow f(0)\leq f(-x^2)$ By $(\beta):$
$\Rightarrow f(-x^2)=f(0)$ By $(\theta):$
$\Rightarrow f(x)=f(0),\forall x\in\mathbb{R}$ $\Rightarrow \boxed{f(x)\equiv c}_\blacksquare$ $\color{blue}\rule{24cm}{0.3pt}$

This post has been edited 1 time. Last edited by hectorleo123, Aug 28, 2023, 9:39 PM

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ismayilzadei1387

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ismayilzadei1387

#6 h Oct 1, 2023, 5:22 PM • 1 Y

Y by FriIzi

f(x)=c with easy substitution

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#7 h Oct 2, 2023, 5:04 PM • 1 Y

Y by pco

ismayilzadei1387 wrote:

f(x)=c with easy substitution

Excellent contribution

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ismayilzadei1387

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ismayilzadei1387

#8 h Oct 2, 2023, 5:13 PM • 1 Y

Y by FriIzi

tadpoleloop wrote:

ismayilzadei1387 wrote:

f(x)=c with easy substitution

Excellent contribution

Hahah i didint want to post same substitution.

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#9 h Mar 19, 2025, 1:37 PM

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Firstly, from $P(x,0)$ , we obtain that for all real numbers $x$ ,
$f(x) \leq f(0).$
Summing $P(x,y)$ and $P(y,x)$ , we derive the inequality
$f(x+y) \leq f(xy).$
Now, define $Q(x,y)$ as follows:
$Q(x,y) \Rightarrow f(x+y) \leq f(xy).$
Substituting $Q(x, -x)$ , we obtain
$f(0) \leq f(-x^2).$
This implies that for all negative real numbers $x$ , we have
$f(x) = f(0).$
Finally, choosing $x, y$ such that $x < 0$ , $y > 0$ , and $x+y < 0$ , and applying $P(x,y)$ , we get
$f(0) \leq f(y)$ for all positive real numbers $y$ .

Thus, we conclude that $f$ is a constant function.

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