1. Let $c$ be a real number. If there exists $x \in[1,2]$, such that $\max \left\{\left|x+\frac{c}{x}\right|,\left|x+\frac{c}{x}+2\right|\right\} \geqslant 5$,
find the range of values for $c$.
(Yang Xiaoming, problem contributor)
1. Note that,
$$
\begin{array}{l}
\max \left\{\left|x+\frac{c}{x}\right|,\left|x+\frac{c}{x}+2\right|\right\} \\
=\left|x+\frac{c}{x}+1\right|+1 .
\end{array}
$$

The given condition can be transformed into: there exists $x \in[1,2]$, such that $\left|x+\frac{c}{x}+1\right| \geqslant 4$.

Let $g(x)=x+\frac{c}{x}+1(x \in[1,2])$.
We will discuss different cases.
(1) If $c=0$, then $g(x)=x+1$.

When $x \in[1,2]$, $|g(x)| \leqslant 30$, then $g(x)>0$.
By the convexity of $g(x)$ on the interval $[1,2]$, we have
$$
\begin{array}{l}
\max _{1 \leqslant x \leqslant 2}|g(x)|=\max _{1 \leqslant x \leqslant 2} g(x) \\
=\max \{g(1), g(2)\} \\
=\max \left\{2+c, 3+\frac{c}{2}\right\} \geqslant 4 .
\end{array}
$$

Solving this, we get $c \geqslant 2$.
(3) If $c<0$, then $g(x)$ is monotonically increasing on the interval $[1,2]$, so
$$
\begin{array}{l}
\max _{1 \leqslant x \leqslant 2}|g(x)|=\max \{|g(1)|,|g(2)|\} \\
=\max \left\{|2+c|,\left|3+\frac{c}{2}\right|\right\} \geqslant 4 .
\end{array}
$$

Since $c<0$, we have
$$
\begin{array}{l}
2+c \leqslant-4 \text { or } 3+\frac{c}{2} \leqslant-4 \\
\Rightarrow c \leqslant-6 .
\end{array}
$$

In summary, $c \in(-\infty,-6] \cup[2,+\infty)$.