mavropnevma wrote: For all triplets $a,b,c$ of (pairwise) distinct real numbers, prove the inequality $$ \left | \dfrac {a+b} {a-b} \right | + \left | \dfrac {b+c} {b-c} \right | + \left | \dfrac {c+a} {c-a} \right | \geq 2$$ and determine all cases of equality. Let $\frac{a+b}{a-b}=x, \frac{b+c}{b-c}=y, \frac{c+a}{c-a}=z$. Notice that $xy+yz+zx=-1$. We have $(\left | \dfrac {a+b} {a-b} \right | + \left | \dfrac {b+c} {b-c} \right | + \left | \dfrac {c+a} {c-a} \right |)^2=x^2+y^2+z^2+2|xy|+2|yz|+2|zx|$ $\geq (x+y+z)^2-2(xy+yz+zx)+2|xy+yz+zx|=4 \Rightarrow \left | \dfrac {a+b} {a-b} \right | + \left | \dfrac {b+c} {b-c} \right | + \left | \dfrac {c+a} {c-a} \right | \geq 2$

mavropnevma wrote: For all triplets $a,b,c$ of (pairwise) distinct real numbers, prove the inequality $$ \left | \dfrac {a+b} {a-b} \right | + \left | \dfrac {b+c} {b-c} \right | + \left | \dfrac {c+a} {c-a} \right | \geq 2$$ Let $\dfrac {a+b} {a-b}=x$, $ \dfrac {b+c} {b-c} =y$ and $\dfrac {c+a} {c-a}=z$. Hence, $(|x|+|y|+|z|)^2=(x+y+z)^2+2+2\sum_{cyc}|xy|\geq2-2\sum_{cyc}xy=4$.

See also here https://artofproblemsolving.com/community/c6h1061424p4596212 http://www.artofproblemsolving.com/community/c6h472766p2702111 https://artofproblemsolving.com/community/c6h614833p3658929

mavropnevma wrote: For all triplets $a,b,c$ of (pairwise) distinct real numbers, prove the inequality $$ \left | \dfrac {a+b} {a-b} \right | + \left | \dfrac {b+c} {b-c} \right | + \left | \dfrac {c+a} {c-a} \right | \geq 2$$and determine all cases of equality. Slovakia TST 2014

mavropnevma wrote: For all triplets $a,b,c$ of (pairwise) distinct real numbers, prove the inequality $$ \left | \dfrac {a+b} {a-b} \right | + \left | \dfrac {b+c} {b-c} \right | + \left | \dfrac {c+a} {c-a} \right | \geq 2$$and determine all cases of equality. Generalization