[tex]\it a(a-1)=b+c|_{:a} \Rightarrow a-1=\dfrac{b+c}{a}|_{+1} \Rightarrow a=\dfrac{a+b+c}{a} \Rightarrow \dfrac{1}{a}=\dfrac{a}{a+b+c}\\ \\ \\ Analog,\ \dfrac{1}{b}=\dfrac{b}{a+b+c},\ \ \dfrac{1}{c}=\dfrac{c}{a+b+c}\\ \\ \\ \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1[/tex]
