Lời giải:
Áp dụng BĐT Bunhiacopxky với các số thực \(a,b,c\)
\(\left(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}\right)\left ( \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \right )\geq \left ( \frac{a}{c}+\frac{b}{a}+\frac{c}{b} \right )^2\)
\(\left(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}\right)\left ( \frac{b}{a}+\frac{c}{b}+\frac{a}{c} \right )\geq \left ( \frac{b}{c}+\frac{c}{a}+\frac{a}{b} \right )^2\)
Cộng hai vế trên thu được:
\(\left(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2}\right)\left(\frac{a+c}{b}+\frac{b+c}{a}+\frac{a+b}{c}\right)\geq \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2+\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)^2\)
Tiếp tục áp dụng Bunhiacopxky:
\([\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2+\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)^2](1+1)\geq \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)^2\)
Suy ra:
\(\left(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2}\right)\left(\frac{a+c}{b}+\frac{b+c}{a}+\frac{a+b}{c}\right)\geq \frac{1}{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)^2\)
\(\Leftrightarrow \frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2}\geq \frac{1}{2}\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right)\)
Ta có đpcm.
Dấu bằng xảy ra khi \(a=b=c\)