Có $\frac{(a+b)^2}{ab}=\frac{a^2+2ab+b^2}{ab}=\frac{a^2+b^2}{2ab}+\frac{a^2+b^2+4ab}{2ab}\geq \frac{a^2+b^2}{2ab}+\frac{2ab+4ab}{2ab}=\frac{1}{2}.(\frac{a}{b}+\frac{b}{a})+3$
Cmtt: $\frac{(b+c)^2}{bc}\geq \frac{1}{2}.(\frac{b}{c}+\frac{c}{b})+3;\frac{(c+a)^2}{ac}\geq \frac{1}{2}.(\frac{c}{a}+\frac{a}{c})+3$
$\Rightarrow \frac{(a+b)^2}{ab}+\frac{(b+c)^2}{bc}+\frac{(c+a)^2}{ca}\geq 9+\frac{1}{2}.((\frac{a}{b}+\frac{a}{c})+(\frac{b}{a}+\frac{b}{c})+(\frac{c}{a}+\frac{c}{b}))$
$\geq 9+\frac{1}{2}.(\frac{4a}{b+c}+\frac{4b}{a+c}+\frac{4c}{b+a})\Rightarrow $ đpcm