Lời giải:
Ta có \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\)
\(\Leftrightarrow \frac{ab^2+bc^2+ca^2}{abc}=\frac{a^2b+b^2c+c^2a}{abc}\)
\(\Leftrightarrow ab^2+bc^2+ca^2=a^2b+b^2c+c^2a\)
\(\Leftrightarrow ab^2+bc^2+ca^2-a^2b-b^2c-c^2a=0\)
\(\Leftrightarrow ab(b-a)+bc(c-b)+ac(a-c)=0\)
\(\Leftrightarrow ab(b-a)-bc[(b-a)+(a-c)]+ac(a-c)=0\)
\(\Leftrightarrow (b-a)(ab-bc)+(a-c)(ac-bc)=0\)
\(\Leftrightarrow b(b-a)(a-c)-c(a-c)(b-a)=0\)
\(\Leftrightarrow (b-a)(a-c)(b-c)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b=a\\a=c\\b=c\end{matrix}\right.\)
Do đó luôn tồn tại hai số bằng nhau (đpcm)
\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{b}\)
\(\Rightarrow\dfrac{a^2c}{abc}+\dfrac{b^2a}{abc}+\dfrac{c^2b}{abc}=\dfrac{b^2c}{abc}+\dfrac{a^2b}{abc}+\dfrac{c^2a}{abc}\)
\(\Rightarrow\dfrac{a^2c+b^2a+c^2b}{abc}=\dfrac{b^2c+a^2b+c^2a}{abc}\)
\(\Rightarrow a^2c+b^2a+c^2b=b^2c+a^2b+c^2a\)
\(\Rightarrow a^2c+b^2a+c^2b-b^2c-a^2b-c^2a=0\)
\(\Rightarrow\left(a^2c-c^2a\right)+\left(b^2a-a^2b\right)+\left(c^2b-b^2c\right)=0\)
\(\Rightarrow ac\left(a-c\right)+ab\left(b-a\right)+bc\left(c-b\right)=0\)
\(\Rightarrow ac\left(a-c\right)+ab\left(b-a\right)+bc\left(c-b+a-a\right)=0\)
\(\Rightarrow ac\left(a-c\right)+ab\left(b-a\right)+bc\left(c-a\right)+bc\left(a-b\right)\)
\(\Rightarrow c\left(a-c\right)\left(a-b\right)+b\left(a-b\right)\left(c-a\right)=0\)
\(\Rightarrow c\left(a-c\right)\left(a-b\right)-b\left(a-b\right)\left(a-c\right)=0\)
\(\Rightarrow\left(c-b\right)\left(a-c\right)\left(a-b\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}c=b\\a=c\\a=b\end{matrix}\right.\)(Tồn tại ít nhất 2 số bằng nhau)
