3. Let the function $f(x)$ satisfy: for any non-negative real number $x$, there is $f(x)=f(1) \cdot x+\frac{f(2)}{x}-1$, then the minimum value of $f(x)$ on $(0,+\infty)$ is $\qquad$ .
$$
\left\{\begin{array} { l } 
{ f ( 1 ) = f ( 1 ) + f ( 2 ) - 1 , } \\
{ f ( 2 ) = 2 f ( 1 ) + \frac { f ( 2 ) } { 2 } - 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
f(1)=\frac{3}{4}, \\
f(2)=1
\end{array} \Rightarrow f(x)=\frac{3}{4} x+\frac{1}{x}-1 \geqslant 2 \sqrt{\frac{3}{4}}-1\right.\right.
$$
$=\sqrt{3}-1$, equality holds when $x=\frac{2 \sqrt{3}}{3}$.