\(VT=\dfrac{1}{\dfrac{a}{b}+\dfrac{c}{a}+1}+\dfrac{1}{\dfrac{b}{c}+\dfrac{a}{b}+1}+\dfrac{1}{\dfrac{c}{a}+\dfrac{b}{c}+1}\)
\(\left(\dfrac{a}{b},\dfrac{b}{c},\dfrac{c}{a}\right)\rightarrow\left(x^3,y^3,z^3\right)\)\(\Rightarrow xyz=1\).
\(VT=\sum\dfrac{1}{x^3+y^3+1}\le\sum\dfrac{1}{xy\left(x+y\right)+xyz}=\sum\dfrac{z}{x+y+z}=1\)
Dấu = xảy ra khi x=y=z=1 hay a=b=c