Example 1.8.3. Suppose that \(a, b, c\) are three positive real numbers satisfying that
\[
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=16
\]

Find the minimum and maximum value of
\[
P=\left(a^{4}+b^{4}+c^{4}\right)\left(\frac{1}{a^{4}}+\frac{1}{b^{4}}+\frac{1}{c^{4}}\right)
\]
Solution. We denote $x, y, m, n$ as in the previous problem and also denote
$$p=\sum_{c y c} \frac{a^{2}}{b^{2}} ; q=\sum_{c y c} \frac{b^{2}}{a^{2}}$$

The expression $P$ can be rewritten as
$$P=3+\sum_{c y c} \frac{a^{4}}{b^{4}}=3+p^{2}+q^{2}-2(p+q) \text {. }$$

The hypothesis yields that $x+y=13$. Moreover, we have
$$\begin{array}{c}
x y=3+m+n(1) ; x^{2}+y^{2}=p+q+2(x+y)(2) \\
p^{2}+q^{2}=\sum_{c y c} \frac{a^{4}}{b^{4}}+2(p+q)(3) ; p q=3+\sum_{c y c} \frac{a^{4}}{b^{2} c^{2}}+\sum_{c y c} \frac{b^{2} c^{2}}{a^{4}}(4) \\
m^{2}+n^{2}=\sum_{c y c} \frac{a^{4}}{b^{2} c^{2}}+\sum_{c y c} \frac{b^{2} c^{2}}{a^{4}}+2(m+n)(5) ; x^{3}+y^{3}=m n+6(m+n)+9(6)
\end{array}$$

The two results (4) and (5) combined show that
$$p q=3+m^{2}+n^{2}-2(m+n)(7)$$

According to (1) and (2) and noticing that $x+y=13$, we have
$$p+q=13^{2}-2.13-2(3+m+n)=137-2(m+n)$$

The equation (6) shows that
$$\begin{array}{c}
13^{3}-39(3+m+n)=m n+6(m+n)+9 \Rightarrow m n=2071-45(m+n) \\
\Rightarrow p q=3+(m+n)^{2}-2 m n-2(m+n)=t^{2}+88 t-4139
\end{array}$$
where $t=m+n$. Therefore $P$ can be expressed as a function of $t$ as
$$\begin{aligned}
P=2+(p+q-1)^{2}-2 p q & =2+4(68-t)^{2}-2\left(t^{2}+88 t-4139\right) \\
& =2 t^{2}-720 t+26776=f(t)
\end{aligned}$$

Taking into account the preceding problem, we obtain $39.25 \geq t=m+n \geq 38$ and the conclusion follows: the minimum of $P$ is $f(39.25)=\frac{12777}{8}$, with equality for $a=b=\frac{(11 \pm \sqrt{105}) c}{4}$ or permutations; the maximum of $P$ is $f(38)=2304$, with equality for $\frac{a}{b}=\frac{b}{c}=\frac{3 \pm \sqrt{5}}{2}$ or permutations.