3. Let $a>0$, the function $f(x)=x+\frac{100}{x}$ has a minimum value of $m_{1}$ on the interval $(0, a]$, and a minimum value of $m_{2}$ on the interval $[a,+\infty)$. If $m_{1} m_{2}=2020$, then the value of $a$ is $\qquad$ .
3.1 or 100.

Notice that, $f(x)$ is monotonically decreasing in the interval $(0,10]$ and monotonically increasing in the interval $[10,+\infty)$.
When $a \in(0,10]$, $m_{1}=f(a), m_{2}=f(10)$;
When $a \in[10,+\infty)$, $m_{1}=f(10), m_{2}=f(a)$.
Therefore, we always have
$$
\begin{array}{l}
f(a) f(10)=m_{1} m_{2}=2020 \\
\Rightarrow a+\frac{100}{a}=\frac{2020}{20}=101 \\
\Rightarrow a=1 \text { or } 100 .
\end{array}
$$