Question

If $a=\frac{x}{y+z}$, $b=\frac{y}{z+x}$ and $c=\frac{z}{x+y}$, then prove that $\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=1$.

Answer 1

Given :

$a=\frac{x}{y+z}$, $b=\frac{y}{z+x}$ and $c=\frac{z}{x+y}$

Consider, $\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=\frac{\frac{x}{y+z}}{1+\frac{x}{y+z}}+\frac{\frac{y}{z+x}}{1+\frac{y}{z+x}}+\frac{\frac{z}{x+y}}{1+\frac{z}{x+y}}$

$=\frac{\frac{x}{y+z}}{\frac{y+z+x}{y+z}}+\frac{\frac{y}{z+x}}{\frac{z+x+y}{z+x}}+\frac{\frac{z}{x+y}}{\frac{x+y+z}{x+y}}$

$=\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}$

$=\frac{x+y+z}{x+y+z}=1$

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