Inequality and function

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Inequality and function

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srnjbr

#1 h Friday at 4:26 PM

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Find all f:R--R such that for all x,y, yf(x)+f(y)>=f(xy)

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ektorasmiliotis

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ektorasmiliotis

#2 h Friday at 5:15 PM

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solution: f(x)=0

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pco

#3 h Friday at 5:31 PM

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ektorasmiliotis wrote:

solution: f(x)=0

Plus a lot lot of others
You should post your proof

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pco

#4 h Friday at 5:32 PM

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srnjbr wrote:

Find all f:R--R such that for all x,y, yf(x)+f(y)>=f(xy)

Let $P(x,y)$ be the assertion $yf(x)+f(y)\ge f(xy)$

Let $x,y\ne 0$ :
Adding $P(\frac xy,y)$ and $P(\frac xy,-y)$ , we get $f(y)+f(-y)\ge f(x)+f(-x)$
And so (swapping $x,y$ ) : $f(x)+f(-x)=c$ for some constant $c$ and $\forall x\ne 0$

$P(x,-y)$ $\implies$ $-yf(x)+c-f(y)\ge c-f(xy)$ and so $f(xy)\ge yf(x)+f(y)$
And so, comparing with $P(x,y)$ $f(xy)=yf(x)+f(y)$ $\forall x,y\ne 0$

Swapping there $x,y$ and subtracting : $(y-1)f(x)=(x-1)f(y)$ $\forall x,y\ne 0$
And so $f(x)=a(x-1)$ for some constant $a$ and $\forall x\ne 0$

Let then $x\ne 0$ : $P(0,x)$ $\implies$ $xf(0)+a(x-1)\ge f(0)$ $\implies$ $(x-1)(f(0)+a)\ge 0$ and so $f(0)=-a$

And so $\boxed{f(x)=a(x-1)\quad\forall x}$ which indeed fits, whatever is $a\in\mathbb R$

This post has been edited 1 time. Last edited by pco, Yesterday at 8:50 AM
Reason: Typo

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srnjbr

#5 h Friday at 6:32 PM • 2 Y

Y by aidan0626, pco

f(0)=-a.
Thank you very much for answering

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pco

#6 h Yesterday at 8:50 AM

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srnjbr wrote:

f(0)=-a.

Indeed

, thanks
Edited

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