Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)(ab+bc+ac)\geq (a+b+c)^2\)
\(\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\geq \left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right)^2\)
Nhân theo vế 2 BĐT trên:
\(\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2(ab+bc+ac).\frac{a+b+c}{abc}\geq [(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})]^2\)
\(\Leftrightarrow \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq [(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})]^2\)
\(\Leftrightarrow \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\geq (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$