Question

$\frac{ {a}^{2} }{b + c} = \frac{ {b}^{2} }{c + a} = \frac{ {c}^{2} }{a + b} = 1 \: prove \: that \: \frac{a}{a + 1} + \frac{b}{b + 1} + \frac{c}{c + 1} = 2$
plz slove

Answer 1

Hello.  I hope this helps you.

Step-by-step explanation:

We are given a² / ( b + c ) = 1, which gives a² = b + c.

So

$\displaystyle\frac{a}{a+1}=\frac{a^2}{a(a+1)}=\frac{a^2}{a+a^2}=\frac{b+c}{a+b+c}$

Similarly,

b / ( b + 1 ) = ( c + a ) / ( a + b + c )

c / ( c + 1 ) = ( a + b ) / ( a + b + c )

Adding these three equations together gives

$\displaystyle\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\\ \\=\frac{(b+c) + (c+a) + (a+b)}{a+b+c} \\ \\= \frac{2(a+b+c)}{a+b+c}\\ \\=2$