3. Let $f(x)$ be a monotonic function defined on the interval $(0,+\infty)$. If for any $x \in(0,+\infty)$, we have
$$
f\left(f(x)-2 \log _{2} x\right)=4 \text {, }
$$

then the solution set of the inequality $f(x)<6$ is $\qquad$
3. $\{x \mid 0<x<4\}$.

From the problem, we know there exists a positive constant $c$, such that $f(c)=4$, and for any $x \in(0,+\infty)$, we have
$$
f(x)-2 \log _{2} x=c \text {. }
$$

When $x=c$, we have
$$
f(c)=2 \log _{2} c+c=4,
$$

By monotonicity, we know this equation has a unique solution $c=2$.
Thus, $f(x)=2 \log _{2} x+2$.
From $f(x)<6 \Rightarrow 2 \log _{2} x+2<6 \Rightarrow 0<x<4$.
Rearranging equation (1) and taking the tangent, we get
$$
\begin{array}{l}
\frac{a-b}{1+a b}=\frac{c+d}{1-c d} . \\
\text { Let } b-a=t \text {. Then } \\
t(c d-1)=\left(a^{2}+a t+1\right)(c+d) . \\
\text { Let } t=a^{2}+1 \text {. Then } \\
c d-1=(a+1)(c+d) \\
\Rightarrow(c-(a+1))(d-(a+1)) \\
\quad=a^{2}+2 a+2 .
\end{array}
$$

Thus, we obtain a set of general solutions:
$$
b=a^{2}+a+1, c=a+2, d=a^{2}+3 a+3,
$$

and $a<c<b<d$.
Below are the constructions for $m=3,5,7$.
$\arctan 1+\arctan 2+\arctan 3=\pi$,
$\arctan 1+\arctan 2+\arctan 4+$
$\arctan 15+\arctan 98=2 \pi$,
$\arctan 1+\arctan 2+\arctan 4+\arctan 15+$
$\arctan 185+\arctan 211+\arctan 16928$ $=3 \pi$,

Subsequently, we can construct recursively, solve for the last arctangent value, add three arctangent values, and use mathematical induction to complete the proof.