if-a-2-b-c-b-2-c-a-c-2-a-b-1-then-find-the-value-of-1-1-a-1-1-b-1-1-c-

Question Number 36978 by jayanta11 last updated on 07/Jun/18

i f \frac{a^{2}}{b + c} = \frac{b^{2}}{c + a} = \frac{c^{2}}{a + b} = 1 t h e n f i n d t h e v a l u e o f

\frac{1}{1 + a} + \frac{1}{1 + b} + \frac{1}{1 + c}

Commented by behi83417@gmail.com last updated on 07/Jun/18

a^{2} = b + c, b^{2} = c + a, c^{2} = a + b

\Rightarrow a^{2} + b^{2} + c^{2} = 2 a + 2 b + 2 c

{(a - 1)}^{2} + {(b - 1)}^{2} + {(c - 1)}^{2} = 3 \Rightarrow

a - 1 = b - 1 = c - 1 = 1 \Rightarrow a = b = c = 2

Σ \frac{1}{1 + a} = 3 \times \frac{1}{3} = 1

o r :

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

\Rightarrow Σ \frac{1}{1 + a} = Σ \frac{a}{a + b + c} = \frac{a + b + c}{a + b + c} = 1 .

Answered by ajfour last updated on 07/Jun/18

a^{2} + a = b^{2} + b = c^{2} + c = a + b + c = s

\Rightarrow a (a + 1) = b (b + 1) = c (c + 1) = s

\frac{1}{1 + a} + \frac{1}{1 + b} + \frac{1}{1 + c} = \frac{a}{s} + \frac{b}{s} + \frac{c}{s} = 1 .

Facebook Tweet Pin