Let-a-b-c-be-the-lengths-of-the-sides-of-a-triangle-Show-that-abc-a-b-c-b-c-a-c-a-b-

Question Number 7883 by 314159 last updated on 23/Sep/16

L e t a, b, c b e t h e l e n g t h s o f t h e s i d e s o f a t r i a n g l e .

S h o w t h a t a b c ⩾ (a + b - c) (b + c - a) (c + a - b) .

Commented by sou1618 last updated on 23/Sep/16

x = a + b - c > 0

y = c + a - b > 0 a, b, c : l e n g t h o f t r i a n g l e

z = b + c - a > 0

x + y ⩾ 2 \sqrt{x y} > 0

y + z ⩾ 2 \sqrt{y z} > 0

z + x ⩾ 2 \sqrt{z x} > 0

(x + y) (y + z) (z + x) ⩾ 8 x y z

(2 a) \times (2 c) \times (2 b) ⩾ 8 (a + b - c) (c + a - b) (b + c - a)

a b c ⩾ (a + b - c) (b + c - a) (c + a - b) .

Commented by sou1618 last updated on 23/Sep/16

w h e n \dots

a b c = (a + b - c) (b + c - a) (c + a - b)

\Leftrightarrow x = y = z

\Leftrightarrow {\begin{cases} a + b - c = b + c - a \\ b + c - a = c + a - b \\ c + a - b = a + b - c \end{cases}

\Leftrightarrow a = b = c

Commented by Rasheed Soomro last updated on 23/Sep/16

N i c e!

Answered by prakash jain last updated on 02/Oct/16

see comments