Question

Answer 1

\(\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{b+c}=\dfrac{1}{2024}\\ \Leftrightarrow\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\left(a+b+c\right)=\dfrac{a+b+c}{2024}\\ \Leftrightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{b+c}=\dfrac{2024}{2024}\\ \Leftrightarrow\dfrac{c}{a+b}+1+\dfrac{b}{a+c}+1+\dfrac{a}{b+c}+1=1\\ \Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=-2\)