If $\frac{(b−c)}{a} + \frac{(a+c)}{b} + \frac{(a−b)}{c}=1$ and $a-b+c \neq 0 $, then prove that $\frac 1a = \frac 1b + \frac 1c$

Question

The question given is

If $\dfrac{(b−c)}{a} + \dfrac{(a+c)}{b} + \dfrac{(a−b)}{c}=1$ and $a-b+c \neq 0 $ then prove that $\dfrac 1a = \dfrac 1b + \dfrac 1c$

I tried to take $abc$ on the right hand side after taking the LCM, but ended up with $b^2(c-a)+a^2(b+c)+c^2(a-b)=abc$. I could not simplify any further. Please provide only hints, not complete solution.

Answer 1

$\dfrac{(b−c)}{a} + \dfrac{(a+c)}{b} + \dfrac{(a−b)}{c}=1$ and $a-b+c \neq 0$

Let $b = a + c - x$ where $x \neq 0$. Then

$\dfrac{(a-x)}{a} + \dfrac{(b + x)}{b} + \dfrac{(x-c)}{c} = 1$

$\left(1 - \dfrac x a \right) + \left(1 + \dfrac x b \right) + \left( \dfrac x c - 1 \right) = 1$

$\dfrac x b + \dfrac x c = \dfrac x a$

$\dfrac 1 b + \dfrac 1 c = \dfrac 1 a$

Answer 2

$$\dfrac{(b−c)}{a} + \dfrac{(a+c)}{b} + \dfrac{(a−b)}{c}=1$$ Add $1$ to $\dfrac{(a−b)}{c}$ and subtract $1$ from $\dfrac{(a+c)}{b}$ We get,

$$\dfrac{(b−c)}{a} + \left[\dfrac{(a+c)}{b}-1\right] + \left[\dfrac{(a−b)}{c}+1\right]=1$$

$$\implies\dfrac{(b−c)}{a} + \dfrac{(a+c-b)}{b} + \dfrac{(a+c−b)}{c}=1$$ $$\implies + \dfrac{(a+c-b)}{b} + \dfrac{(a+c−b)}{c}=1-\dfrac{(b−c)}{a}$$ $$\implies \dfrac{(a+c-b)}{b} + \dfrac{(a+c−b)}{c}=\dfrac{(a+c-b)}{a}$$ $$\implies \dfrac{1}{b} + \dfrac{1}{c}=\dfrac{1}{a}$$

Yo!