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


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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 byprakash jain last updated on 18/May/17

$S = a^{2} + b^{2} + c^{2} + 2 a b c$ $4 - S = {(a + b + c)}^{2} - S$ $= 2 a b + 2 b c + 2 c a - 2 a b c$ $= 2 (a b + b c + c a - a b c) . . . (A)$ $a < b + c \Rightarrow 2 a < a + b + c \Rightarrow a < 1 \Rightarrow (1 - a) > 0$ $\Rightarrow a < 1 a l s o b < 1, c < 1$ $(1 - a) (1 - b) (1 - c) > 0$ $(1 - a - b + a b) (1 - c) > 0$ $1 - a - b + a b - c + a c + b c - a b c > 0$ $1 - (a + b + c) + (a b + b c + c a - a b c) > 0$ $- 1 + (a b + b c + c a - a b c) > 0$ $\Rightarrow (a b + b c + c a - a b c) > 1 . . . (B)$ $s u b s t i t u t i n g B i n A$ $4 - S > 2 \Rightarrow 2 > S ◼$
Commented byRasheedSindhi last updated on 18/May/17

$X cellent!!!$
Commented byprakash jain last updated on 18/May/17

$Thanks$
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