Example 7.2.6 Let $a, b, c$ be three positive real numbers, and satisfy $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=16$, find the minimum and maximum values of the following expression.
$$P=\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \text { (Pham Kim Hung) }$$
First, we find the minimum value. For this, assume $a \geq b \geq c$, so we have
$$\sum_{c y c} \frac{a}{b}-\sum_{c y c} \frac{b}{a}=\frac{(a-b)(a-c)(c-b)}{a b c} \leq 0$$

Let $x=\frac{a}{b} \geq 1, y=\frac{b}{c} \geq 1$, then the given condition becomes $x+y+\frac{1}{x y}+\frac{1}{x}+\frac{1}{y}+x y=13$

Let $x+y=s, x y=t$, then $P=s+\frac{1}{t}$, and $s+t+\frac{s}{t}+\frac{1}{t}=13 \Rightarrow s=\frac{13 t-t^{2}-1}{t+1}$,

Therefore, $P=f(t)=\frac{13 t-t^{2}-1}{t+1}+\frac{1}{t}$, then
$$f^{\prime}(t)=\frac{(13-2 t)(t+1)-\left(13 t-t^{2}-1\right)}{(t+1)^{2}}-\frac{1}{t^{2}}=\frac{15}{(t+1)^{2}}-\frac{t^{2}+1}{t^{2}}$$

We easily get $f^{\prime}(t)=0 \Leftrightarrow\left(t^{2}+1\right)(t+1)^{2}=15 t^{2} \Leftrightarrow\left(t+\frac{1}{t}+1\right)^{2}=16 \Leftrightarrow t+\frac{1}{t}=3$ Since $t=a b \geq 1, f^{\prime}(t)=0 \Leftrightarrow t=t_{0}=\frac{3+\sqrt{5}}{2}$, and by the assumption $x, y \geq 1$, we have
$t+1-s=(x-1)(y-1) \geq 0$ i.e., $t+1 \geq s$, so
$\frac{13 t-t^{2}-1}{t+1} \leq t+1 \Leftrightarrow 2 t^{2}-11 t+2 \geq 0 \Rightarrow t \geq \frac{11+\sqrt{105}}{4}>t_{0}$, thus $f(t)$ is a strictly decreasing function. To find the minimum value of $f(t)$, we only need to find the maximum value of $t$. Notice that $s^{2} \geq(x+y)^{2} \geq 4 t$, so we have
$$\begin{array}{l}
\left(13 t-t^{2}-1\right)^{2} \geq 4 t(t+1)^{2} \Rightarrow\left(13-t-\frac{1}{t}\right)^{2} \geq 4\left(t+\frac{1}{t}+2\right) \\
\Rightarrow\left(t+\frac{1}{t}\right)^{2}-30\left(t+\frac{1}{t}\right)+161 \geq 0 \Leftrightarrow\left(t+\frac{1}{t}-7\right)\left(t+\frac{1}{t}-23\right) \geq 0
\end{array}$$

Moreover, $t+\frac{1}{t}<23$ (since $1 \leq t \leq 13$), so we must have $t+\frac{1}{t} \geq 7$ or
$$t^{2}-7 t+1 \leq 0 \Leftrightarrow t \leq \frac{7+3 \sqrt{5}}{2}$$

For $t=\frac{7+\sqrt{5}}{2}$, we have $s=2 \sqrt{t}=\sqrt{14+6 \sqrt{5}}=3+\sqrt{5}$, thus
$\min f(t)=s+\frac{1}{t}=3+\sqrt{5}+\frac{2}{7+3 \sqrt{5}}=\frac{13-\sqrt{5}}{2}$, equality holds when $\frac{a}{b}=\frac{b}{c}=\sqrt{t}=\frac{3+\sqrt{5}}{2}$ or its permutations.
Similarly, using the same method, we have $\max f(t)=\frac{13+\sqrt{5}}{2}$, equality holds when $\frac{a}{b}=\frac{b}{c}=\frac{3-\sqrt{5}}{2}$ or its permutations.