Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that $$ (x+y)f(2yf(x)+f(y))=x^{3}f(yf(x)), \ \ \ \forall x,y\in \mathbb{R}^{+}.$$ (Albania)
Problem
Source: Balkan BMO Shortlist 2015 A4
Tags: functional equation, algebra, function
05.08.2019 18:45
parmenides51 wrote: Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that $$ (x+y)f(2yf(x)+f(y))=x^{3}f(yf(x)), \ \ \ \forall x,y\in \mathbb{R}^{+}.$$ (Albania) Obviously $f(x)$ is injective. Then just set $x=2$ and $y=6$ and you get a contradiction with the fact that $f(x)>0$ $\forall x>0$ Hence $\boxed{\text{No such function}}$
01.10.2019 18:32
pco wrote: parmenides51 wrote: Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that $$ (x+y)f(2yf(x)+f(y))=x^{3}f(yf(x)), \ \ \ \forall x,y\in \mathbb{R}^{+}.$$ (Albania) Obviously $f(x)$ is injective. Then just set $x=2$ and $y=6$ and you get a contradiction with the fact that $f(x)>0$ $\forall x>0$ Hence $\boxed{\text{No such function}}$ How do you know that that function is injective?
01.10.2019 20:07
ImBadAtGeometry wrote: How do you know that that function is injective? If $f(a)=f(b)$, then comparing $P(a,y)$ with $P(b,y)$, we get : $\frac{a^3}{a+y}=\frac{b^3}{b+y}$ $\forall y>0$ Which obviously implies $a=b$
21.10.2020 13:00
$f$ is injective (can be shown by plugging in $P(a, x)$ and $P(b, x)$, where $f(a) = f(b)$), so plugging $P(\sqrt2, \sqrt2)$ yields $1 = -\sqrt2$, hence there are no solutions.
25.06.2021 19:22
If $f(a)=f(b)$ for $a>b$, then $P\left(a,\frac{ab^3+a^3b}{a^3-b^3}\right)-P\left(b,\frac{ab^3+a^3b}{a^3-b^3}\right)$ gives a contradiction, hence $f$ is injective. $P(2,6)\Rightarrow f(12f(2)+f(6))=f(6f(2))\Rightarrow 6f(2)+f(6)=0$, so no solutions exist.
10.10.2021 04:12
The answer is $\boxed{\text{no such function}}$. Let $P(x,y)$ denote the given assertion. Claim: $f$ is injective. Proof: Suppose $f(a)=f(b)$. $P(a,y): (a+y)f(2yf(a)+f(y))=a^{3}f(yf(a))$ $P(b,y): (b+y)f(2yf(a)+f(y))=b^{3}f(yf(a))$ Since we won't be dividing by zero, dividing the two equations gives $\frac{a+y}{b+y}=\frac{a^3}{b^3}$, so $\frac{a^3}{a+y}=\frac{b^3}{b+y}$, for all $y>0$, which implies that $a=b$. $P(x,x^3-x): x^3f(2(x^3-x)f(x)+f(x^3-x))=x^3f((x^3-x)f(x))\implies 2(x^3-x)f(x)+f(x^3-x)=(x^3-x)f(x)$, so $(x^3-x)f(x)>2(x^3-x)f(x)$, a contradiction.
28.05.2023 23:55
obviously f is injective then p(x,x) we get 2f( (2x+1) f(x) )=x^2 f( x f(x) ) then plug x by Sqrt[2] and f is injective so we find f(sqrt2)=0 which is a contradiction because f(x)>0
28.05.2023 23:56
so there is no solutions of f
10.10.2023 13:36
The function $f$ is injective And take $x=2,y=6$ $6f(2)+f(6)=0$ contradicition.
23.05.2024 22:26
Hehe funny, assume there exists such function. from $\frac{f(2yf(x)+f(y))}{f(yf(x))}=\frac{x^3}{x+y}$ we obtain that $f$ is injective. Plugging $y=x^3-x, x>1$ in yields: $(x+x^3-x)f(2(x^3-x)f(x)+f(x^3-x))=x^3f((x^3-x)f(x)) \Leftrightarrow f(2(x^3-x)f(x)+f(x^3-x))=f((x^3-x)f(x))$ Keeping in mind that $f$ is injective we get: $2(x^3-x)f(x)+f(x^3-x)=(x^3-x)f(x) \Leftrightarrow (x^3-x)f(x)+f(x^3-x)=0$, contradicting the assumption. Hence, there are no such functions.
18.10.2024 08:39
Who in the world ever decided to make such a troll question No such $f$ exists. $f$ is injective since if you take $y=1$ and if $f(a)=f(b)$ for $a,b\in\mathbb{R}^+$, then $f(2f(a)+f(1))=f(2f(b)+f(1))=\frac{a^3}{a+1}f(f(a))=\frac{b^3}{b+1}f(f(b))$ and so since $f(x)=\frac{x^3}{x+1}$ is strictly increasing for $x>0$ $a=b$. Now remembering the FE battlecry: David Yang was claimed to have wrote: DURR WE WANT STUFF TO CANCEL We set $x+y=x^3$, I.e. $x^3-x=y$ for $x>1$. Unfortunately we then have $f(2yf(x)+f(y))=f(yf(x))$ or $2yf(x)+f(y)=yf(x)$, i.e. $yf(x)=-f(y)<0$, contradiction. To motivate the sub $y=x^3-x$, notice we have $f$ injective so we might want to do something about the two oddly different terms $2yf(x)+f(y)$ and $yf(x)$ because we can eliminate $f$ now. But then we want the $f$s of both sides equal, i.e. $x+y=x^3$.