\(B=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\)
Áp dụng bất đẳng thức Cauchy: \(x+y\ge2\sqrt{xy}\) (Dấu bằng xảy ra khi a=b), ta có: \(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}=2\)
Tương tự: \(\frac{b}{c}+\frac{c}{b}\ge2\), \(\frac{a}{c}+\frac{c}{a}\ge2\)
Suy ra \(B\ge2+2+2=6\)
Vậy GTNN của B là 6 khi \(\left\{\begin{matrix}\frac{a}{b}=\frac{b}{a}\\\frac{b}{c}=\frac{c}{b}\\\frac{c}{a}=\frac{a}{c}\end{matrix}\right.\Leftrightarrow a=b=c\)
Áp dụng BĐT Cauchy ta có :
ab+ba≥2;bc+cb+ac+ca≥2ab+ba≥2;bc+cb+ac+ca≥2
Cộng vế →B≥6
Áp dụng bđt AM-GM cho 2 số dương ta có:
\(B=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)
\(\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}+2\sqrt{\frac{b}{c}.\frac{c}{b}}+2\sqrt{\frac{a}{c}.\frac{c}{a}}=2.3=8\)
Dấu "=" xảy ra khi a = b = c
