Question

If a:b = b:c then prove that 1 upon 1 + a + 1 upon 1 + b + 1 upon 1 + c = 1​

Answer 1

Answer:

$\frac{(1 + a)(1 + b) + (1 + b)(1 + c) + (1 +a)(1 + c) }{(1 + a)(1 + b)(1 + c)} = 1 \\ \\ hence \: \: proved.$

Step-by-step explanation:

$\frac{a}{b} = \frac{b}{c} = \gamma \\ \\ \\ then \: a = \gamma \times b \\ \\ \\ b = \gamma c \\ \\ a= { \gamma }^{2} c \\ \\ \\ lhs = \frac{1}{1 + a} + \frac{1}{1 + b} + \frac{1}{1 + c} = \frac{(1 + a)(1 + b) + (1 + b)(1 + c) + (1 +a)(1 + c) }{(1 + a)(1 + b)(1 + c)} \\ \\ = \frac{ 1 + ab + a + b + 1 + b + c + bc + 1 + a + c + ac}{(1 + a)(1 + b)(1 + c)} \\ \\ = \frac{3 + 2a + 2b + 2c + ab + bc + ca}{(1 + a)(1 + b)(1 + c)} \\ \\ = \frac{3 + 2 { \gamma }^{2} c + 2 \gamma c + 2c + { \gamma }^{2}c \gamma + \gamma c + c { \gamma }^{2}c }{(1 + { \gamma }^{2} c)(1 + \gamma c)(1 + c)} \\ \\ = \frac{3 + 2 { \gamma }^{2} c + 2 \gamma c + 2c + { \gamma }^{2}c \gamma + \gamma c + c { \gamma }^{2}c }{3 + 2 { \gamma }^{2} c + 2 \gamma c + 2c + { \gamma }^{2}c \gamma + \gamma c + c { \gamma }^{2}c } \\ \\ = 1 \\ \\ = rhs$