# Exercice 2

• Determiner toutes les fonctions $f:]0,+\infty[ \rightarrow ]0,+\infty [$ telles que
$f(x)\cdot f(y)=f(y)f(xf(y))+\dfrac{1}{xy}$
pour tout $x,y \gt0$

• Since no one posted a solution , here's one :

Let $P(x,y)$ be the assertion $f(x)f(y)=f(y)f(xf(y))+\frac 1{xy}$

$P(1,x)$ :$f(f(x))=f(1)-\frac 1{xf(x)}$

$P(f(x),y)$ :$f(f(x))=f(f(x)f(y))+\frac 1{f(x)f(y)y}$

Subtracting, we get $f(f(x)f(y))=f(1)-\frac 1{xf(x)}-\frac 1{f(x)f(y)y}$
Swapping $x,y$ and subtracting, we get :

$\frac 1{xf(x)}+\frac 1{f(x)f(y)y}=\frac 1{yf(y)}+\frac 1{f(x)f(y)x}$

Which is $x(f(x)-1)=y(f(y)-1)$ and so $f(x)=1+\frac ax$ for some $a$

Plugging this back in original equation, we get $a=1$

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