Find the minimum of the value $\max\left(\frac{1}{ac}+b,\frac{1}{a}+bc,\frac{a}{b}+c\right)$ Let $a,b,c>0$ find the following minimum of the value
$$\max\left(\dfrac{1}{ac}+b,\dfrac{1}{a}+bc,\dfrac{a}{b}+c\right)$$
the book is hint:when $a=b=c=1$,then the minimum of the value is $2$

Let $M =\max \left[ \frac{1}{ac}+b, \frac{1}{a}+bc, \frac{a}{b}+c \right]$

Case I: $c\ge 1$
\begin{align*}
  M &=\max \left[ \frac{1}{a}+bc, \frac{a}{b}+c \right]
      \tag{$\frac{1}{a}+bc \ge \frac{1}{ac}+b$} \\
    & \ge \max \left[ 2\sqrt{\frac{bc}{a}}, 2\sqrt{\frac{ac}{b}} \right]
      \tag{$AM \ge GM$} \\
    & \ge \max \left[ 2\sqrt{\frac{b}{a}}, 2\sqrt{\frac{a}{b}} \right]
      \tag{$c \ge 1$} \\
    & \ge 2
\end{align*}

The lower bound is achievable when $a=b=c=1$


Case II: $c<1$
\begin{align*}
  M &=\max \left[ \frac{1}{ac}+b, \frac{a}{b}+c \right]
      \tag{$\frac{1}{ac}+b > \frac{1}{a}+bc$} \\
    & \ge \max \left[ 2\sqrt{\frac{b}{ac}}, 2\sqrt{\frac{ac}{b}} \right]
      \tag{$AM \ge GM$} \\
    & \ge 2
\end{align*}
When $ac=b$,
\begin{align*}
  \frac{b}{a} & < 1 \\
  \frac{a}{b}+\frac{b}{a} &> 2 \\
  \frac{1}{b}+b & \ge 2 \\
  M &=\max \left[ \frac{1}{b}+b, \frac{a}{b}+\frac{b}{a} \right] \\
    &> 2
\end{align*}

No values of $c\in (0,1)$ give $M=2$


The smallest possible $M$ is $2$ when $a=b=c=1$.
