Giải thích các bước giải:
Ta có:
$P=\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}$
$\to P\ge 3\sqrt{\dfrac{a^2}{b^2}\cdot\dfrac{b^2}{c^2}\cdot\dfrac{c^2}{a^2}}$
$\to P\ge 3$
$\to 2P=\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)+\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)$
$\to 2P\ge \dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}+3$
$\to 2P\ge \left(\dfrac{a^2}{b^2}+1\right)+\left(\dfrac{b^2}{c^2}+1\right)+\left(\dfrac{c^2}{a^2}+1\right)$
$\to 2P\ge \dfrac{2a}{b}+\dfrac{2b}{c}+\dfrac{2c}{a}$
$\to P\ge \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}$
