Lời giải:
Mặc định đk $a,b,c\neq 0$
Áp dụng BĐT Cô-si cho các số dương ta có:
\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\geq 2\sqrt{\frac{a^2}{b^2}.\frac{b^2}{c^2}}=2|\frac{a}{c}|\geq \frac{2a}{c}\)
\(\frac{a^2}{b^2}+\frac{c^2}{a^2}\geq 2\sqrt{\frac{a^2}{b^2}.\frac{c^2}{a^2}}=2|\frac{c}{b}|\geq \frac{2c}{b}\)
\(\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq 2\sqrt{\frac{b^2}{c^2}.\frac{c^2}{a^2}}=2|\frac{b}{a}|\geq \frac{2b}{a}\)
Cộng theo vế:
\(2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\geq 2\left(\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\right)\)
\(\Leftrightarrow \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq \frac{c}{b}+\frac{b}{a}+\frac{a}{c}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c\)
