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).


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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

$w h e n . . .$ $a b c = (a + b - c) (b + c - a) (c + a - b)$ $\Leftrightarrow x = y = z$ $\Leftrightarrow {\begin{cases} a + b - c = b + c - a \\ b + c - a = c + a - b \\ c + a - b = a + b - c \end{cases}$ $\Leftrightarrow a = b = c$
Commented by Rasheed Soomro last updated on 23/Sep/16

$N i c e!$
	Answered by prakash jain last updated on 02/Oct/16

	$see comments$
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