If-a-b-and-c-are-the-sides-of-a-triangle-and-a-b-c-2-then-prove-that-a-2-b-2-c-2-2abc-lt-2-

Question Number 13294 by Tinkutara last updated on 17/May/17

If a, b and c are the sides of a triangle

and a + b + c = 2, then prove that

a^{2} + b^{2} + c^{2} + 2 a b c < 2

Commented by prakash jain last updated on 18/May/17

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

4 - S = {(a + b + c)}^{2} - S

= 2 a b + 2 b c + 2 c a - 2 a b c

= 2 (a b + b c + c a - a b c) \dots (A)

a < b + c \Rightarrow 2 a < a + b + c \Rightarrow a < 1 \Rightarrow (1 - a) > 0

\Rightarrow a < 1 a l s o b < 1, c < 1

(1 - a) (1 - b) (1 - c) > 0

(1 - a - b + a b) (1 - c) > 0

1 - a - b + a b - c + a c + b c - a b c > 0

1 - (a + b + c) + (a b + b c + c a - a b c) > 0

- 1 + (a b + b c + c a - a b c) > 0

\Rightarrow (a b + b c + c a - a b c) > 1 \dots (B)

s u b s t i t u t i n g B i n A

4 - S > 2 \Rightarrow 2 > S

Commented by RasheedSindhi last updated on 18/May/17

X cellent!!!

Commented by prakash jain last updated on 18/May/17

Thanks