abc of maths

Add $\displaystyle \frac{b+c}{b+c} + \frac{c+a}{c+a} + \frac{a+b}{a+b} = 3$ to the both sides we get

$\displaystyle \frac{a+b+c}{b+c} + \frac{a+b+c}{c+a} + \frac{a+b+c}{a+b} = 3$

$\displaystyle 5\left(\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b}\right) = 3$

$\displaystyle \left(\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b}\right) = \boxed{\displaystyle \frac{3}{5}}$ ~~~

This is my first solution, I hope you enjoy it!

Let's wiggle with $\frac{a}{b+c}$ . First, because $a=5-(b+c)$ ,

$\frac{a}{b+c} = \frac{5-(b+c)}{b+c} = \frac{5}{b+c} - 1$ .......(1)

Analogously,

$\frac{b}{a+c} = \frac{5}{a+c} - 1$ ........................(2)

$\frac{c}{a+b} = \frac{5}{a+b} - 1$ ........................(3)

Adding (1), (2) and (3), because the LHS equals zero, we get

$0 = 5(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}) - 3 \rightarrow (\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}) = \frac{3}{5}$

Then, $3+5 = \boxed{8}$ , the answer.

mine

$(a+b+c)(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+3$

$5(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})=0+3$

$\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}=\frac{3}{5}$

Add 3 on both sides and distribute the three uniformly to all terms. Then take a+b+c common and solve the problem.