-1
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$f(a,b,c)$=$\frac{a-b}{a-c}$ + $\frac{b-c}{b-a}$ + $\frac{c-a}{c-b}$

$f(a,b,c)$ = $\frac{(a-b)*(b-a)*(c-b) + (a-c)*(b-c)*(c-b) + (c-a)*(a-c)*(b-a)}{(a-c)*(b-a)*(c-b)}$

$f(a,b,c)$ = $\frac{(a-b)^2*(b-c) + (c-a)*(b-c)^2 + (a-c)^2*(a-b)}{(a-c)*(b-a)*(c-b)}$

$f(a,b,c)$ = $\frac{(a-b)^2*(b-c) + (c-a)*(b-c)^2 + (a-c)^2*(a-b)}{(a-c)*(b-a)*(c-b)}$

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1
  • $\begingroup$ The title matches not the body. Per chance you could revise it? Thanks. $\endgroup$
    – awllower
    Jan 23, 2013 at 10:55

1 Answer 1

Reset to default
3
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Substituting $1$ for $a$, $2$ for $b$, and $3$ for $c$ yields $\frac{1}{2}-\frac{2}{2}+\frac{4}{2}=\frac{3}{2}$.

So, no.

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