Hello all,
I need some help on this problem:
If $\displaystyle \frac{y+z}{pb+qc} = \frac{z+x}{pc+qa} = \frac{x+y}{pa+qb} $, then show that $\displaystyle \frac{2(x+y+z)}{a+b+c} = \frac{(b+c)x + (c+a)y + (a+b)z}{bc+ ca + ab} $
I tried ...
The sum of $\displaystyle \frac{y+z}{pb+qc} = \frac{z+x}{pc+qa} =\frac{x+y}{pa+qb} = \frac{2(x+y+z)}{(p+q)(a+b+c)} $
It is close to the proof except for $\displaystyle (p+b) $ in the denominator. What should I do next? It can disappear if I assume $\displaystyle (p+b) = 1 $
$\displaystyle \frac{y+z}{pq+bc}=\frac{ay+az}{pab+qac}$
$\displaystyle \frac{z+x}{pc+qa}=\frac{bz+bx}{pbc+qab}$
$\displaystyle \frac{x+y}{pa+qb}=\frac{cx+cy}{pac+qbc}$
add thse three equations
$\displaystyle \frac{y+z}{pq+bc}+\frac{z+x}{pc+qa}+\frac{x+y}{pa+ qb}=\frac{ay+az}{pab+qac}+\frac{bz+bx}{pbc+qab}+\f rac{cx+cy}{pac+qbc}$