Proving that $ f(1)=\frac{1-\sqrt{5}}{2}$ for this function 
Let $f:(0,+\infty)\mapsto R$ be a strictly increasing function  such that $\forall x\ge0,$ $$f(x)+\frac{1}{x}\ge0,  \qquad f(x)f\left(f(x)+\frac{1}{x}\right)=1.$$ 
  Show that $$f(1)=\frac{1-\sqrt{5}}{2}.$$ 

Please give an example that satisfies these conditions.
Thanks in advance.

$f(x)f(f(x)+\frac{1}{x})=1$
$x=1 $ gives $ f(1)f(f(1)+1)=1$
$x=f(1)+1\ge0$ gives $ f(f(1)+1)f(f(f(1)+1)+\frac{1}{f(1)+1})=1$
By replacing $f(f(1)+1)$ by $\frac{1}{f(1)}$ you get $\frac{1}{f(1)}f(\frac{1}{f(1)}+\frac{1}{f(1)+1})=1$
Multiply both sides by $f(1)$ to get $f(\frac{1}{f(1)}+\frac{1}{f(1)+1})=f(1)$
Then you know that the function is strictly increasing and therefore injective.
So $\frac{1}{f(1)}+\frac{1}{f(1)+1}=1$
And then solve for $f(1)$.
Let $x=f(1)$
$\frac{1}{x}+\frac{1}{x+1}=1$
Multiply by $x$ and $x+1$ to get $x+1+x=x(x+1)$
$2x+1=x^2+x$
$0=x^2-x-1$
$x=\cfrac{1\pm\sqrt{5}}{2}$
