Ta có:
\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+1=\dfrac{a^2}{ab}+\dfrac{b^2}{bc}+\dfrac{c^2}{ca}+\dfrac{b^2}{b^2}\)
\(\ge\dfrac{\left(a+2b+c\right)^2}{ab+bc+ca+b^2}=\dfrac{\left(a+b\right)^2+2\left(a+b\right)\left(b+c\right)+\left(b+c\right)^2}{\left(a+b\right)\left(b+c\right)}\)
\(=\dfrac{a+b}{b+c}+\dfrac{b+c}{a+b}+2\)
Sorry bác Neet tới đây e bí mất 
Ta có: \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge3.\sqrt[3]{\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{a}}=3\)(1)
\(\dfrac{a+b}{b+c}+\dfrac{b+c}{a+b}\ge2.\sqrt{\dfrac{a+b}{b+c}.\dfrac{b+c}{a+b}}=2\)
\(\Leftrightarrow\dfrac{a+b}{b+c}+\dfrac{b+c}{a+b}+1\ge3\)(2)
Từ (1), (2), ta có: \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}-\dfrac{a+b}{b+c}-\dfrac{b+c}{a+b}-1\ge0\)
\(\Leftrightarrow\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{a+b}{b+c}+\dfrac{b+c}{a+b}+1\)
Dấu "=" xảy ra khi \(a=b=c\)
