Example 1.8.1. Suppose that $a, b, c$ are three positive real numbers satisfying
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=13 .$$

Find the minimum value of
$$P=\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) .$$
SOLUTION. Let now
$$x=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}, y=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$$
then we obtain that
$$x^{2}=2 y+\sum_{c y c} \frac{a^{2}}{b^{2}} ; y^{2}=2 x+\sum_{c y c} \frac{b^{2}}{a^{2}}$$

Because $x+y=10$, AM-GM inequality yields that
$$P-3=x^{2}+y^{2}-2(x+y) \geq \frac{1}{2}(x+y)^{2}-2(x+y)=50-20=30$$
therefore $P \geq 33$ and $\min P=33$ with equality for
$$x=y \Leftrightarrow \sum_{c y c} \frac{a}{b}=\sum_{c y c} \frac{b}{a} \Leftrightarrow(a-b)(b-c)(c-a)=0,$$
combined with the hypothesis $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=10$, we conclude that $P=33$ if and only if $(a, b, c) \sim(2 \pm \sqrt{3}, 1,1)$ and permutations.