BĐT cần chứng minh <=> \(\left(a+b+c\right)\left(\frac{b}{a^2}+\frac{c}{b^2}+\frac{a}{c^2}\right)\ge9\)
Áp dụng BĐT Cauchy 3 số ta có: \(\left(a+b+c\right)\left(\frac{b}{a^2}+\frac{c}{b^2}+\frac{a}{c^2}\right)\ge3\sqrt[3]{abc}.3.\sqrt[3]{\frac{b}{a^2}.\frac{c}{b^2}.\frac{a}{c^2}}=9.\sqrt[3]{abc}.\sqrt[3]{\frac{1}{abc}}=9\)
Dấu "=" xảy ra khi a = b = c