Cho abc ≠ 1 hoặc abc ≠ –1 và $\frac{ab+1}{b}=\frac{bc+1}{c}=\frac{ca+1}{a}$  .

Chứng minh rằng a = b = c.

Ta có: $\frac{ab+1}{b}=\frac{bc+1}{c}=\frac{ca+1}{a}$

$\iff a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}$

$\iff \left\{\begin{array}{l}a-b=\frac{1}{c}-\frac{1}{b}\\ b-c=\frac{1}{a}-\frac{1}{c}\\ c-a=\frac{1}{b}-\frac{1}{a}\end{array}\right.$

$\iff \left\{\begin{array}{l}a-b=\frac{b-c}{bc}\\ b-c=\frac{c-a}{ac}\\ c-a=\frac{a-b}{ab}\end{array}\right.$

$\iff \left(a-b\right)\left(b-c\right)\left(c-a\right)=\frac{b-c}{bc}.\frac{c-a}{ac}.\frac{a-b}{ab}$

$\iff \left(a-b\right)\left(b-c\right)\left(c-a\right)-\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{a^2{b}^2{c}^2}=0$

$\iff \left(a-b\right)\left(b-c\right)\left(c-a\right)\left(1-\frac{1}{a^2{b}^2{c}^2}\right)=0$

$\iff \left(a-b\right)\left(b-c\right)\left(c-a\right)=0$ (vì abc ≠ 1, abc ≠ –1)

$\iff \left[\begin{array}{l}a-b=0\\ b-c=0\\ c-a=0\end{array}\right.$

$\iff \left[\begin{array}{l}a=b\\ b=c\\ c=a\end{array}\right.$

• Nếu a = b thì $\frac{1}{a}-\frac{1}{c}=0$ . Do đó a = c, suy ra a = b = c;

• Nếu b = c thì $\frac{1}{c}-\frac{1}{b}=0$ . Do đó b = c, suy ra a = b = c;

• Nếu c = a thì $\frac{1}{b}-\frac{1}{a}=0$  . Do đó b = a, suy ra a = b = c.

Vậy a = b = c.